ML Aggarwal Class 11 Maths Solutions Chapter 07 Linear Inequalities

PERMUTATIONS AND COMBINATIONS

Introduction

In daily life, we come across many problems of finding the number of ways of arranging or selecting objects.

Suppose you have a suitcase with a number lock. The number lock has three wheels each marked with 10 digits from 0 to 9. The lock can be opened if 3 specific digits are arranged in a particular order with no repetition of digits. Somehow, you have forgotten this particular order of the digits. The question arises - "How many arrangements you have to check with to open the lock?" To answer this question, you may immediately start listing all the possible arrangements of 10 digits taken 3 at a time with no repetitions. But, this method of listing the arrangements is very tedious and time consuming, because the number of possible arrangements is large.

In the present chapter, we will learn some basic techniques of counting which will enable us to answer the above question without actually listing 3-digit arrangements. In fact, these techniques will be useful in determining the number of different ways of arranging and selecting objects in a wide variety of situations.

7.1 Fundamental Principle of Counting

Let us first consider the following examples.

Example 1. Sania has 2 school bags and 3 water bottles. In how many different ways can she carry these objects to her school choosing one each?
Answer: Since there are 2 school bags, a school bag can be chosen in 2 different ways. Similarly, a water bottle can be chosen in 3 different ways. For every choice of a school bag, there are three choices of a water bottle. Therefore, the number of different ways in which a pair of a school bag and a water bottle can be chosen is 2 × 3 = 6. Hence Sania can carry these objects to her school choosing one each in 6 different ways.

If we represent the 2 school bags by S₁, S₂ and the 3 water bottles by W₁, W₂, W₃, then the six different ways can be shown by the following tree diagram:


In simple words: There are 2 school bags to pick. For each one, you can pick any of 3 water bottles. Multiply 2 times 3 to get 6 total ways.

Exam Tip: Always multiply the number of choices for each step - this is the fundamental counting principle. When the choices for each step are independent, multiply them together.

Example 2. Numbers 1, 2, 3 and 4 are written on four cards. How many 2-digit numbers can be formed by placing two cards side by side?
Answer: You may pick any one of the four cards and put it down representing the digit as ten's place. After this, pick any card out of the remaining 3 cards and put it to the right of the previous card representing a digit at unit's place. For every choice of a card at ten's place, there are 3 choices for a card at unit's place. Therefore, the number of different numbers which can be formed by placing two cards side by side is 4 × 3 = 12.


In simple words: Pick any of 4 cards for the first place. Then pick any of the 3 left over cards for the second place. This gives 4 times 3 equals 12 numbers.

Exam Tip: When selecting without repetition, the number of choices decreases at each step. Always multiply the number of choices available at each position.

Fundamental Principle of Counting

If an event can occur in m different ways, and if when it has occurred, a second event can occur in n different ways, then the total number of different ways of occurrence of the two events is m × n.

The above principle is also called the multiplication principle of counting.

In Example 1, the required number of ways of carrying a school bag and a water bottle was the number of ways of the occurrence of the following events in succession:

(i) the event of choosing a school bag
(ii) the event of choosing a water bottle.

The above principle of counting can be generalised to any finite number of events. Now, let us consider the following example:

Gopal has 8 shirts, 5 pants and 2 pairs of shoes. In how many different ways can he dress up himself with?

Since Gopal has 8 shirts, he can choose a shirt in 8 ways. Similarly, he can choose a pant in 5 ways. For every choice of a shirt, there are 5 choices of a pant. Therefore, a shirt and a pant can be chosen in 8 × 5 ways - that is, 40 ways. Further, since he has 2 pairs of shoes, he can choose a pair of shoes in 2 ways. For every choice of a shirt and a pant, there are 2 choices of a pair of shoes. Therefore, a shirt, a pant and a pair of shoes (complete dress) can be chosen in 40 × 2 ways - that is, 80 ways. Hence Gopal can dress up himself in 80 different ways.

The above example leads to the following generalisation of the principle of counting:

If an event can occur in m different ways, and if when it has occurred, a second event can occur in n different ways, following which a third event can occur in p different ways and so forth, then the total number of different ways of occurrence of all the events is m × n × p ......

Illustrative Examples

Example 1. (i) You can go from Delhi to Jaipur either by car or by bus or by train or by air. In how many ways can you plan your journey from Delhi to Jaipur and back to Delhi?
(ii) Suppose you like variety and you don't want to return by the same mode of transport. How many different ways are possible?
Answer:
(i) You can go from Delhi to Jaipur in 4 different ways, and you can return from Jaipur to Delhi in 4 different ways. So, by multiplication principle of counting, there are 4 × 4 = 16 different ways of going from Delhi to Jaipur and returning to Delhi.

(ii) You can go from Delhi to Jaipur in 4 different ways. Corresponding to each of these, there are 3 different ways of returning from Jaipur to Delhi (since a different mode of transport has to be chosen while returning). Hence, the required number of ways is 4 × 3 = 12.
In simple words: For the first part, you have 4 choices going and 4 choices returning, giving 16 total ways. For the second part, after picking a way to go, only 3 ways remain for returning, giving 12 total ways.

Exam Tip: When there is a restriction (different transport), the choices available at the second step reduce. Always account for such restrictions when calculating the total.

Example 2. (i) A lady wants to select one cotton saree and one polyester saree from a textile shop. If there are 10 cotton varieties and 12 polyester varieties, in how many ways can she choose the two sarees?
(ii) If in above case, the lady has limited budget, in how many ways can she choose one saree?
Answer:
(i) The lady can select one cotton saree out of 10 cotton varieties in 10 ways since any of 10 varieties can be selected. Corresponding to each selection of a cotton saree, she can choose a polyester saree in 12 ways. Hence the two sarees (one cotton and one polyester), by multiplication principle of counting, can be selected in 10 × 12 = 120 ways.

(ii) Since only one saree is to be chosen and 10 + 12 = 22 different varieties of sarees are available, one saree (either of cotton or of polyester) can be chosen in 22 ways.
In simple words: When you must pick both types, multiply the options. When you pick only one type out of the total available, add the options together.

Exam Tip: Use multiplication when selecting from multiple categories together. Use addition when selecting one item from multiple options within a single category.

Example 3. In a class, there are 27 boys and 14 girls. In how many ways can the teacher form a team of one boy and one girl from amongst the students of the class to represent the class for a function?
Answer: The teacher can select one boy out of 27 boys in 27 ways since any of 27 boys can be selected. Corresponding to each selection of a boy, the teacher can select one girl in 14 ways since any of 14 girls can be selected. Hence a team consisting of one boy and one girl, by multiplication principle of counting, can be formed in 27 × 14 = 378 ways.
In simple words: Pick any of 27 boys. For each boy, pick any of 14 girls. This gives 27 times 14 equals 378 total pairs.

Exam Tip: For forming a team with one member from each category, multiply the number of choices in each category. The total outcome is the product.

Example 4. Three persons enter a railway carriage, where there are 5 vacant seats. In how many ways can they seat themselves?
Answer: The first man can sit on any of 5 vacant seats. Then the second can sit on any of 4 vacant seats left. And the third can sit on any of 3 vacant seats left. Hence by fundamental principle of counting, the required number of ways is 5 × 4 × 3 = 60.
In simple words: The first person sits in one of 5 seats. The second sits in one of 4 left over seats. The third sits in one of 3 left over seats. Multiply: 5 times 4 times 3 equals 60.

Exam Tip: When the number of available positions decreases with each selection, always account for this reduction. Do not use the same number of choices for all positions.

Example 5. How many words can be formed out of the letters of the word 'MAGIC' taking all the letters at a time (no letter being repeated)?
Answer: There are 5 different letters with which 5 vacant places are to be filled up. The first place can be filled in 5 ways, as any one of the five letters M, A, G, I, C can be placed there. Having filled up the first place in any of the 5 ways, 4 letters are left and any one of them can be placed in second place. Hence the first two places can together be filled up in 5 × 4 ways. Now three letters are left and any of them can be put in third place. After that, 2 letters are left and any of them can be put in fourth place. After that only 1 letter is left and it has to be placed in fifth place. Therefore, the total number of ways of filling up five places is 5 × 4 × 3 × 2 × 1 = 120. Hence, the required number of words formed is 120.
In simple words: Fill the first spot with any of 5 letters. Fill the second spot with any of 4 left over letters. Keep going: third gets 3 choices, fourth gets 2 choices, fifth gets 1 choice. Multiply all: 5 times 4 times 3 times 2 times 1 equals 120.

Exam Tip: When arranging all items with no repetition, the number of arrangements equals 5 times 4 times 3 times 2 times 1. This is called a factorial.

Example 6. How many 4 letter code words are possible using the first 10 letters of the English alphabet if
(i) no letter can be repeated? (NCERT)
(ii) letters may be repeated?
Answer:
(i) No letter can be repeated.
There are 10 different ways for choosing the first letter of the code, 9 different ways for choosing the second letter of the code, 8 ways for choosing the third letter and 7 ways for choosing the fourth letter of the code. Therefore, the total number of 4 letter code words is 10 × 9 × 8 × 7 = 5040.

(ii) Letters can be repeated.
There are 10 different ways for choosing the first letter of the code. As the letters can be repeated, there are 10 different ways for choosing the second letter of the code. Similarly, there are 10 ways of choosing for each of the third and the fourth letter of the code. Therefore, the total number of 4 letter code words is 10 × 10 × 10 × 10 = 10000.
In simple words: Without repeating: each position has fewer choices. First gets 10, second gets 9, third gets 8, fourth gets 7. Multiply to get 5040. With repeating: each position has 10 choices always. Multiply 10 four times to get 10000.

Exam Tip: When repetition is allowed, use the same number of choices for each position. When not allowed, reduce the choices at each step.

Example 7. How many automobile license plates can be made if each plate contains two different letters followed by three different digits?
Answer: The number of ways of forming license plates containing two different letters followed by three different digits is the same as filling five vacant places, first two places by two different letters and the next three places by three different digits. The first place can be filled in 26 ways by any of 26 letters and the second place can be filled in 25 ways by any of 25 remaining letters. The third place can be filled in 10 ways by any of the digits from 0 to 9, the fourth place can be filled in 9 ways by any of the remaining 9 digits and the fifth place can be filled in 8 ways by any of the remaining 8 digits. By multiplication principle of counting, the total number of ways of filling these places - that is, the total number of ways of forming license plates is 26 × 25 × 10 × 9 × 8 = 468000.
In simple words: Pick a letter (26 ways), then another different letter (25 ways). Pick a digit (10 ways), then another different digit (9 ways), then a third different digit (8 ways). Multiply all: 26 times 25 times 10 times 9 times 8 equals 468000.

Exam Tip: Track which positions must be different from prior choices. Each restriction removes one option from the available pool for that position.

Example 8. In a certain city, all telephone numbers have six digits, the first two digits always being 41 or 42 or 46 or 62 or 64. How many telephone numbers have all six digits distinct?
Answer: The number of telephone numbers of six digits having all six distinct digits and the first two digits being 41 or 42 or 46 or 62 or 64 is the same as filling six vacant places, first two places by any of the numbers 41, 42, 46, 62 or 64 and the remaining four places by the other digits, of course, all six digits distinct. We note that the first two places (that is, I and II) together can be filled in 5 ways. As two different digits are already used in first place, so we are left with only 8 different digits. Third place can be filled in 8 ways by any of 8 remaining digits. Similarly, fourth place can be filled in 7 ways, fifth in 6 ways and sixth place in 5 ways. By multiplication principle of counting, the number of ways of filling all the six places (under given condition) is 5 × 8 × 7 × 6 × 5. Therefore, the required number of telephone numbers is 5 × 8 × 7 × 6 × 5 = 8400.
In simple words: The first two digits are fixed in one of 5 ways. This uses 2 digits, leaving 8 digits. Now fill positions 3 through 6 using 8, 7, 6, and 5 choices. Multiply: 5 times 8 times 7 times 6 times 5 equals 8400.

Exam Tip: When initial digits are pre-set by rule, count how many digits remain, then apply the multiplication principle to the remaining positions.

Example 9. How many 3-digit even numbers can be formed from the digits 1, 2, 3, 4, 5, 6 if the digits can be repeated?
Answer: There are six different digits with which 3 vacant places are to be filled up satisfying the given conditions. As the required 3-digit numbers are even, we start filling in unit's place and the options for this place are 2, 4, and 6 only. So unit's place can be filled up in 3 different ways. Having filled up unit's place, the ten's place can be filled up by any of the six digits 1, 2, 3, 4, 5 and 6 in six different ways as the digits can be repeated. Similarly, the hundred's place can be filled up in 6 ways. Therefore, by the multiplication principle of counting, the required number of 3-digit even numbers is 3 × 6 × 6 = 108.
In simple words: Start with the units place - only 2, 4, or 6 make it even, so 3 choices. The tens place can be any digit, so 6 choices. The hundreds place can be any digit, so 6 choices. Multiply: 3 times 6 times 6 equals 108.

Exam Tip: For even/odd numbers, fill the units (or ones) place first based on the parity restriction, then fill remaining places. This simplifies the counting.

Example 10. How many numbers lying between 100 and 1000 can be formed with the digits 0, 1, 2, 3, 4, 5 if the repetition of digits is not allowed?
Answer: The numbers lying between 100 and 1000 consist of 3 digits. There are six different digits with which three vacant places are to be filled up, repetition of digits not allowed. As the required numbers are of 3-digits, the first place (hundred's place) has to be non-zero so the options for this place are 1, 2, 3, 4, 5. So hundred's place can be filled up in 5 different ways. Having filled up hundred's place, the second place (ten's place) can be filled up in 5 ways with any one of the remaining 5 digits as the repetition of digits is not allowed. The third place (unit's place) can be filled up in 4 ways with any one of the remaining 4 digits. By fundamental principle of counting, the required number of numbers lying between 100 and 1000 is 5 × 5 × 4 = 100.
In simple words: The first digit cannot be 0, so pick from 1, 2, 3, 4, 5 - that's 5 ways. The second digit can be any of the 5 remaining digits (including 0 now). The third digit can be any of the 4 remaining. Multiply: 5 times 5 times 4 equals 100.

Exam Tip: When the first digit cannot be zero, count your choices carefully at each step. Always account for the zero being unavailable at the start but available later.

Example 11. How many 4-digit numbers are there with no digit repeated?
Answer: There are 10 different digits (0 to 9) with which four vacant places are to be filled up, repetition of digits not allowed. As the required numbers are of 4-digits, the first place (thousand's place) has to be non-zero so there are 9 choices. So the first place can be filled up in 9 ways. Having filled up the first place, the second place (hundred's place) can be filled up in 9 ways by any of the remaining 9 digits as the repetition of digits is not allowed. Similarly, the third place (ten's place) can be filled up in 8 ways and the fourth place (unit's place) can be filled up in 7 ways. By fundamental principle of counting, the required number of 4-digit numbers is 9 × 9 × 8 × 7 = 4536.
In simple words: The first digit is 1 through 9 (not 0), giving 9 choices. The second digit is any of 9 remaining digits including 0. The third digit is any of 8 remaining. The fourth is any of 7 remaining. Multiply: 9 times 9 times 8 times 7 equals 4536.

Exam Tip: After filling the first position with a non-zero digit, that digit is removed from the pool, but 0 becomes available. Track the pool size at each step.

Example 12. Find the number of 3-digit odd numbers, when repetition of digits is allowed.
Answer: There are 10 different digits (0 to 9) with which three vacant places are to be filled up satisfying the given conditions, repetition of digits is allowed. As the required 3-digit numbers are odd, we start filling in unit's place and the options for this place are 1, 3, 5, 7 and 9 only. So unit's place can be filled up in 5 different ways. Having filled up unit's place, the ten's place can be filled up by any of the 10 digits (0 to 9) in 10 ways. As the hundred's place has to be non-zero, so this place can be filled up in 9 different ways. By fundamental principle of counting, the required number of 3-digit odd numbers is 5 × 10 × 9 = 450.
In simple words: Start with units place to ensure the number is odd - pick from 1, 3, 5, 7, 9, giving 5 choices. The tens place can be any digit 0 through 9, giving 10 choices. The hundreds place cannot be 0, so pick from 1 through 9, giving 9 choices. Multiply: 5 times 10 times 9 equals 450.

Exam Tip: Start with the units place when you need to ensure odd or even. Start with the leftmost (hundreds) place when you need to ensure it is not zero. Apply whichever restriction is more constraining first.

Example 13. Find the number of 4-digit odd numbers, when repetition of digits is not allowed.
Answer: There are 10 different digits (0 to 9) with which four vacant places are to be filled up satisfying the given conditions, repetition of digits is not allowed. As the required 4-digit numbers are odd, we start filling in unit's place and the options for this place are 1, 3, 5, 7 and 9 only. So unit's place can be filled up in 5 different ways. After filling unit's place, we fill thousand's place. As the thousand's place has to be non-zero and the repetition of digits is not allowed, so this place can be filled by the remaining 8 digits (one digit is used for unit's place and 0 cannot be filled at this place) in 8 different ways. After filling unit's place and thousand's place, we fill up the remaining two places - that is, hundred's place and ten's place. As two digits are already used and repetition of digits is not allowed, therefore, the hundred's place can be filled up by the remaining 8 digits in 8 ways and the ten's place can be filled up by the remaining 7 digits in 7 ways. By fundamental principle of counting, the required number of 4-digit odd numbers when repetition of digits not allowed is 5 × 8 × 8 × 7 = 2240.
In simple words: Fill the units place with an odd digit - 5 choices. Fill the thousands place with a non-zero remaining digit - 8 choices (one is used, and 0 cannot go here). Fill the hundreds place with any of 8 remaining digits. Fill the tens place with any of 7 remaining. Multiply: 5 times 8 times 8 times 7 equals 2240.

Exam Tip: When you have two restrictions (odd AND no zero in thousands place), fill the most constrained position first, then adjust the choices for the next constrained position accordingly.

Example 14. Find the number of positive integers greater than 6000 and less than 7000 which are divisible by 5, provided that no digit is to be repeated.
Answer: We are to find the number of positive integers between 6000 and 7000 which are divisible by 5. So, we are to fill up four vacant places by 10 different digits (0 to 9) satisfying the given conditions and repetition of digits is not allowed. The thousand's place has to be filled up by the digit 6, so there is only one way to fill up this place. As the numbers are divisible by 5, so the unit's place has to be filled up either by 5 or by 0. Thus, there are only 2 ways to fill up unit's place. Now, the hundred's place can be filled up in 8 ways by any of the remaining 8 digits because at thousand's place and at unit's place two digits have already been used. Similarly, the ten's place can be filled up in 7 ways. Therefore, the number of required positive integers is 1 × 8 × 7 × 2 = 112.
In simple words: The thousands place must be 6 - that's 1 way. For divisibility by 5, the units place is 0 or 5 - that's 2 ways. The hundreds place gets any of 8 remaining digits. The tens place gets any of 7 remaining. Multiply: 1 times 8 times 7 times 2 equals 112.

Exam Tip: For divisibility rules, the units place is restricted first. For range restrictions (like 6000 to 7000), fix the leading digit first, then handle the divisibility rule.

Example 15. In how many ways can 5 persons sit in a car, 2 including the driver in the front seat and 3 in the back seat, if 2 particular persons do not know driving?
Answer: Let us mark the 5 seats by the letters A, B, C, D and E with A as driver's seat. Since 2 particular persons out of the 5 do not know driving, there are 3 choices for seat A, 4 choices for seat B, 3 choices for seat C, 2 choices for seat D and one choice for seat E. Therefore, the total number of seating arrangements is 3 × 4 × 3 × 2 × 1 = 72.
In simple words: The driver's seat needs someone who can drive - 3 choices (out of 5 people, 2 cannot drive). Seat B next to driver - 4 people left, all can sit. Seat C in back - 3 people left. Seat D - 2 left. Seat E - 1 left. Multiply: 3 times 4 times 3 times 2 times 1 equals 72.

Exam Tip: When certain people cannot fill certain roles, restrict the choices for those positions. Start with the most constrained position.

Example 16. Given 5 flags of different colours. How many different signals can be generated by hoisting the flags on a vertical pole (one below the other) if each signal requires the use of
(i) two flags? (NCERT)
(ii) at least two flags? (NCERT)
Answer:
(i) There will be as many different signals as the different number of ways of filling in two vacant places by 5 flags of different colours. The upper vacant place can be filled in 5 different ways by any one of the 5 flags. Having filled up the upper place in any one of the 5 ways, 4 flags are left. The lower place can be filled in by any one of the remaining 4 flags. Therefore, by multiplication principle of counting, the number of different signals that can be generated by hoisting two flags (one below the other) is 5 × 4 = 20.

(ii) As a signal consists of at least 2 flags, so a signal can consist of either 2 flags, 3 flags, 4 flags or 5 flags. Let us count the possible number of signals generated by hoisting 2 flags, 3 flags, 4 flags and 5 flags separately (one below the other).

The number of different signals generated by hoisting 2 flags is 5 × 4 = 20. (see (i))

Similarly, the number of different signals generated by hoisting 3 flags is 5 × 4 × 3 = 60.

The number of different signals generated by hoisting 4 flags is 5 × 4 × 3 × 2 = 120.

The number of different signals generated by hoisting 5 flags is 5 × 4 × 3 × 2 × 1 = 120.

Therefore, the number of different signals generated by hoisting at least two flags is 20 + 60 + 120 + 120 = 320.
In simple words: For part (i), pick one flag for the top spot (5 ways), then pick a different flag for below (4 ways) - multiply to get 20. For part (ii), signals can have 2, 3, 4, or 5 flags. Find the number of signals for each size: 20, 60, 120, and 120. Add them: 20 plus 60 plus 120 plus 120 equals 320.

Exam Tip: For "at least n" problems, break into cases and add the results. For each case, apply the multiplication principle to count arrangements.

Example 17. How many numbers can be formed from the digits 1, 2, 3, 9 if the repetition of digits is not allowed?
Answer: We have to consider 1-digit numbers, 2-digit numbers, 3-digit numbers and 4-digit numbers. Here we are given four digits 1, 2, 3 and 9; repetition of digits not allowed.

Number of 1-digit numbers = 4.

Let us consider 2-digit numbers. The first place (ten's place) can be filled in 4 ways, then second place (unit's place) can be filled in 3 ways. Therefore, number of 2-digit numbers is 4 × 3 = 12.

Similarly, number of 3-digit numbers = 4 × 3 × 2 = 24.

The number of 4-digit numbers = 4 × 3 × 2 × 1 = 24.

Hence, the required number of numbers = 4 + 12 + 24 + 24 = 64.
In simple words: Count numbers by size. One digit: 4 ways. Two digits: 4 times 3 equals 12 ways. Three digits: 4 times 3 times 2 equals 24 ways. Four digits: 4 times 3 times 2 times 1 equals 24 ways. Add all: 4 plus 12 plus 24 plus 24 equals 64.

Exam Tip: When forming numbers of varying lengths from a fixed set of digits, organize by size. Calculate arrangements for each size, then add the totals.

Example 18. How many numbers are there between 100 and 1000 (including 100 but excluding 1000) such that
(i) every digit is either 2 or 5
(ii) there is no restriction
(iii) no digit is repeated
(iv) the digit in hundred's place is 5
(v) the digit in unit's place is 5
(vi) at least one of the digits is 5
(vii) at least one of the digits is repeated
(viii) exactly one digit is 5?
Answer:
(i) Since every digit is 5 or 2, there are 2 ways of filling up of each of three digits (places). Thus the three digits (places) can be filled in 2 × 2 × 2 = 8 ways. Hence required number of numbers = 8.

(ii) The first digit (hundred's place) has to be non-zero so there are 9 choices. Second and third digits can be any of ten digits 0 to 9. Hence required number of numbers = 9 × 10 × 10 = 900 (including 100 but excluding 1000).

(iii) First digit (hundred's place) can be any digit from 1 to 9; second digit can be then chosen in 9 ways, and third digit in 8 ways. Thus required number of numbers with no digit repeated is 9 × 9 × 8 = 648.

(iv) Digit in hundreds place is fixed (5). The other two places can each be filled in 10 ways. Hence the possible ways are 1 × 10 × 10 = 100.

(v) The digit in hundred's place can be filled in 9 ways (1 to 9); the digit in tens place can be filled in 10 ways (0 to 9), and digit in units place can be filled in one way only as it is fixed (5). Hence the required number of numbers = 9 × 10 × 1 = 90.

(vi) First, let us find all numbers between 100 and 1000 which don't have digit 5 in any place. The number of such numbers is 8 × 9 × 9 = 648 (including 100). We have also seen in part (ii) that there are 900 numbers between 100 and 1000 (including 100). Hence there are 900 - 648 = 252 numbers between 100 and 1000 which have at least one digit as 5.

(vii) As there are 900 numbers between 100 and 1000 (including 100) and 648 of them have no digit repeated, so there are 900 - 648 = 252 numbers (including 100) which have at least one digit repeated.

(viii) The numbers with exactly one digit as 5 are:
(a) of the type 5 × × = 1 × 9 × 9 = 81 (where × is a non-five digit)
(b) of the type × 5 × = 8 × 1 × 9 = 72 (first digit is non-zero non-five)
(c) of the type × × 5 = 8 × 9 × 1 = 72.
Thus, there are 81 + 72 + 72 = 225 numbers between 100 and 1000 which have exactly one digit as 5.
In simple words: For part (i), each place is 2 or 5 - multiply 2 × 2 × 2 to get 8. For part (ii), no restrictions - first digit is 1-9 (9 ways), others are 0-9 (10 ways each) - get 900. For part (iii), no repeats - count decreasing choices: 9 × 9 × 8 = 648. For parts (iv) and (v), fix one place and count the rest. For part (vi), use complement - total minus those without 5 = 252. For part (viii), count 5 in each position separately, then add.

Exam Tip: For "at least" problems, often use the complement (total minus the opposite). For "exactly" problems, break into non-overlapping cases by position.

Example 19. How many of the natural numbers from 1 to 1000 have none of their digits repeated?
Answer: We have to consider natural numbers consisting of 1-digit, 2-digits and 3-digits, and find how many of these have distinct digits. Obviously, we are ignoring 1000 as it has 0 as repeated digit.

Number of 1-digit natural numbers = 9 (as it can be any one of 1, 2, 3, …, 9).

Let us consider 2-digit numbers. The ten's place has to be non-zero, so there are 9 choices. As no digit is to be repeated, there are 9 choices for unit's place. Hence the number of 2-digit numbers is 9 × 9 = 81.

Similarly, the number of 3-digit numbers = 9 × 9 × 8 = 648.

Therefore, the number of numbers from 1 to 1000 which have none of their digits repeated is 9 + 81 + 648 = 738.
In simple words: One digit: any 1 through 9 - 9 numbers. Two digits: first is 1-9 (9 ways), second is 0-9 except what was used (9 ways) - 81 numbers. Three digits: first is 1-9 (9 ways), second is 0-9 except what was used (9 ways), third is 0-9 except what was used (8 ways) - 648 numbers. Total: 9 plus 81 plus 648 equals 738.

Exam Tip: When calculating numbers with distinct digits, account for zero's special role - it cannot be the leading digit but is available elsewhere. Adjust choices accordingly at each step.

Example 20. Find the total of natural numbers that can be formed by using the digits 0, 1, 2, 3, 4 and 5, repetition of digits is not allowed.
Answer: We are required to form natural numbers consisting of 1-digit, 2-digits, 3-digits, 4-digits, 5-digits and 6-digits from the digits 0, 1, 2, 3, 4 and 5, without repetition of digits.

Number of 1-digit natural numbers = 5.

Let us form 2-digit natural numbers from the given digits. The ten's place has to be non-zero, so there are 5 choices (1, 2, 3, 4, 5) for this place. As no digit is to be repeated, there are 5 choices for unit's place. Therefore, the number of 2-digit natural numbers is 5 × 5 = 25.

Similarly, the number of 3-digit natural numbers = 5 × 5 × 4 = 100.

The number of 4-digit natural numbers = 5 × 5 × 4 × 3 = 300.

The number of 5-digit natural numbers = 5 × 5 × 4 × 3 × 2 = 600.

The number of 6-digit natural numbers = 5 × 5 × 4 × 3 × 2 × 1 = 600.

Therefore, the number of required natural numbers is 5 + 25 + 100 + 300 + 600 + 600 = 1630.
In simple words: One digit: 1-5 only (5 numbers). Two digits: first from 1-5 (5 ways), second from remaining including 0 (5 ways) - 25 numbers. Keep building: three digits is 100, four digits is 300, five digits is 600, six digits is 600. Total: 5 plus 25 plus 100 plus 300 plus 600 plus 600 equals 1630.

Exam Tip: When zero is available and you form numbers of all possible lengths, zero cannot be the leading digit but becomes available for other positions once the leading digit is chosen. Adjust the pool size accordingly.

Example 21. How many odd numbers less than 1000 can be formed using the digits 0, 1, 4, 5, 7, 8 if the repetition of digits is allowed?
Answer: The numbers less than 1000 will consist of 1-digit, 2-digits or 3-digits. The digits to be used are 0, 1, 4, 5, 7, 8 satisfying the given condition and repetition of digits is allowed.

1-digit odd numbers

Number of 1-digit odd numbers = 3 (1, 5 or 7)

2-digit odd numbers

The number of ways of filling up unit's place = 3 (1, 5 or 7)

As the ten's place has to be non-zero, the number of ways of filling up ten's place = 5 (1, 4, 5, 7 or 8).

Therefore, the number of 2-digit odd numbers = 3 × 5 = 15.

3-digit odd numbers

The number of ways of filling up unit's place = 3 (1, 5 or 7)

The number of ways of filling up ten's place = 6 (0, 1, 4, 5, 7 or 8).

The number of ways of filling up hundred's place = 5 (1, 4, 5, 7 or 8).

Therefore, the number of 3-digit odd numbers = 3 × 6 × 5 = 90.

Therefore, the total number of the required numbers = 3 + 15 + 90 = 108.
In simple words: One digit odd: 1, 5, or 7 - 3 numbers. Two digits odd: units place is 1, 5, or 7 (3 ways), tens place is non-zero from the set (5 ways) - 15 numbers. Three digits odd: units place is 1, 5, or 7 (3 ways), tens place is any digit (6 ways), hundreds place is non-zero (5 ways) - 90 numbers. Total: 3 plus 15 plus 90 equals 108.

Exam Tip: Always fill the units place first when ensuring odd or even, to minimize restrictions on other positions. This often simplifies the counting.

Example 22. How many odd numbers less than 10000 can be formed using the digits 0, 1, 4, 5, 7, 8 if the repetition of digits is not allowed?
Answer: The numbers less than 10000 will consist of 1-digit, 2-digits, 3-digits or 4-digits. The digits to be used are 0, 1, 4, 5, 7, 8 satisfying the given condition and repetition of digits not allowed.

1-digit odd numbers

Number of 1-digit odd numbers = 3 (1, 5 or 7)

2-digit odd numbers

The number of ways of filling up unit's place = 3 (1, 5 or 7).

As the ten's place has to non-zero and repetition of digits not allowed, the number of ways of filling up ten's place = 4 (because one non-zero digit is already used at unit's place).

Therefore, the number of 2-digit odd numbers = 3 × 4 = 12.

3-digit odd numbers

The number of ways of filling up units place = 3 (1, 5 or 7).

After filling unit's place, we fill hundred's place. As the hundred's place has to be non-zero and repetition of digits not allowed, the number of ways of filling hundred's place = 4.

The number of ways of filling up ten's place = 4.

Therefore, the number of 3-digit odd numbers = 3 × 4 × 4 = 48.

4-digit odd numbers

The number of ways of filling up unit's place = 3 (1, 5 or 7).

After filling unit's place, we fill thousand's place. As this place has to be non-zero and repetition of digits not allowed, the number of ways of filling up thousand's place = 4.

The number of ways of filling up hundred's place = 4.

The number of ways of filling up ten's place = 3.

Therefore, the number of 4-digit odd numbers = 3 × 4 × 4 × 3 = 144.

Therefore, the total number of required numbers = 3 + 12 + 48 + 144 = 207.
In simple words: One digit: 1, 5, or 7 - 3 numbers. Two digits: units is 1, 5, or 7 (3 ways), tens is non-zero remaining (4 ways) - 12 numbers. Three digits: units is 1, 5, or 7 (3 ways), hundreds is non-zero remaining (4 ways), tens is any remaining (4 ways) - 48 numbers. Four digits: units is 1, 5, or 7 (3 ways), thousands is non-zero remaining (4 ways), hundreds is any remaining (4 ways), tens is any remaining (3 ways) - 144 numbers. Total: 3 plus 12 plus 48 plus 144 equals 207.

Exam Tip: When no repetition is allowed, track carefully which digits are already used. After placing a digit, reduce the available pool for the next position.

Example 23. In how many ways can 5 different balls be distributed among 3 boxes?
Answer: Each ball can be put into any one of the three boxes in 3 different ways. Therefore, by fundamental principle of counting, the number of ways of distributing 5 different balls among three boxes is 3 × 3 × 3 × 3 × 3 = 3^5 = 243.
In simple words: Each of the 5 balls can go into any of 3 boxes. That is 3 choices per ball. Since there are 5 balls, multiply 3 five times: 3 times 3 times 3 times 3 times 3 equals 243.

Exam Tip: When each object has the same number of independent choices (like boxes to place items), raise that number to the power of how many objects you have.

Example 24. Find the total number of ways of answering 6 multiple choice questions, each question having 4 choices.
Answer: Each question can be answered in 4 ways. Therefore, by fundamental principle of counting, the number of ways of answering 6 multiple choice questions is 4 × 4 × 4 × 4 × 4 × 4 = 4^6.
In simple words: Each of the 6 questions has 4 answer choices. You pick one choice per question. Multiply 4 six times: 4^6.

Exam Tip: When all items have the same number of choices and are independent, use exponentiation: (number of choices) raised to the power (number of items).

Exercise 7.1

Very Short Answer Type Questions (1 to 20):

Question 1. In a cricket match, how many choices can be made for man of the match?
Answer: If we assume the award goes to one player from a standard team, and teams typically have 11 players, then there are 11 different choices for man of the match.
In simple words: Any one of the team players can be chosen. If a team has 11 players, there are 11 choices.

Exam Tip: The exact number depends on context - it could be 11 (one team), 22 (both teams), or another number. Always state your assumption if not given.

Question 2. In a class there are 22 girls and 17 boys. The teacher wants to select either a girl or a boy to represent the class in a function. In how many ways can the teacher make this selection?
Answer: Since the teacher is selecting either a girl or a boy (not both), this is a case of choosing one from one of two groups. The teacher can select a girl in 22 ways or a boy in 17 ways. By the addition principle, the total number of ways is 22 + 17 = 39.
In simple words: Pick a girl (22 ways) OR pick a boy (17 ways). Add the options: 22 plus 17 equals 39.

Exam Tip: Use addition when choosing from mutually exclusive options (either-or). Use multiplication when choosing from independent groups (and).

Question 3. In a class there are 17 girls and 22 boys. In how many ways can the teacher form a team of one girl and one boy from amongst the students of the class to represent the school in a quiz competition?
Answer: The teacher can select one girl out of 17 girls in 17 ways and one boy out of 22 boys in 22 ways. Since both selections happen together, by the multiplication principle of counting, the total number of ways to form a team of one girl and one boy is 17 × 22 = 374.
In simple words: Pick any of 17 girls. Pick any of 22 boys. Multiply: 17 times 22 equals 374.

Exam Tip: When forming a pair or team requiring one member from each group, multiply the sizes of the groups.

Question 4. Of 11 cricket players, one is to be chosen as captain and another as vice captain. How many choices are there?
Answer: The captain can be chosen in 11 ways from the 11 players. Once the captain is chosen, the vice captain can be chosen from the remaining 10 players in 10 ways. By the multiplication principle of counting, the total number of ways is 11 × 10 = 110.
In simple words: Pick a captain from 11 players (11 ways). Pick a vice captain from the 10 remaining players (10 ways). Multiply: 11 times 10 equals 110.

Exam Tip: When selecting two people for distinct roles without repetition, multiply the number of choices: (first role choices) times (second role choices after one is removed).

Question 5. Lata wants to go abroad by air and return by ship. She has a choice of 6 different airlines to go and 4 different ships to return. In how many ways she can perform her journey?
Answer: Lata can choose an airline in 6 different ways. For each choice of airline, she can choose a ship to return in 4 different ways. By the multiplication principle of counting, the total number of ways she can perform her journey is 6 × 4 = 24.
In simple words: Pick one of 6 airlines to go (6 ways). Pick one of 4 ships to return (4 ways). Multiply: 6 times 4 equals 24.

Exam Tip: For two sequential, independent choices (going and returning by different modes), multiply the number of options for each leg.

Question 6. There are 4 doors in Lotus temple. In how many ways can a person enter the temple and leave by a different door?
Answer: A person can enter the temple through any of the 4 doors in 4 ways. To leave by a different door, the person must choose one of the remaining 3 doors. By the multiplication principle of counting, the total number of ways is 4 × 3 = 12.
In simple words: Enter through one of 4 doors (4 ways). Leave through a different door from the remaining 3 (3 ways). Multiply: 4 times 3 equals 12.

Exam Tip: When a choice must differ from a prior choice, reduce the available options for the later choice.

Question 7. Four persons A, B, C and D are to give lectures to an audience. In how many ways can the organiser arrange the order of their presentation?
Answer: The first position can be filled by any of the 4 persons in 4 ways. The second position can be filled by any of the remaining 3 persons in 3 ways. The third position can be filled by any of the remaining 2 persons in 2 ways. The fourth position is filled by the last remaining person in 1 way. By the multiplication principle of counting, the total number of ways is 4 × 3 × 2 × 1 = 24.
In simple words: Pick one of 4 people to go first. Pick one of 3 left for second. Pick one of 2 left for third. The last person goes fourth. Multiply: 4 times 3 times 2 times 1 equals 24.

Exam Tip: When arranging all n distinct items in order (permutation), the total is n × (n-1) × (n-2) × ... × 1.

Question 8. Find the number of 4 letter words, with or without meaning, which can be formed out of the letters of the word ROSE (repetition of letters not allowed).
Answer: The word ROSE has 4 distinct letters. To form 4-letter words using all these letters with no repetition, the first position can be filled in 4 ways by any of the 4 letters. The second position can be filled in 3 ways by any of the remaining 3 letters. The third position can be filled in 2 ways. The fourth position can be filled in 1 way. By the multiplication principle of counting, the total number of 4-letter words is 4 × 3 × 2 × 1 = 24.
In simple words: Arrange the 4 letters R, O, S, E in all possible orders. Choose first letter (4 ways), second letter (3 ways), third letter (2 ways), fourth letter (1 way). Multiply: 4 times 3 times 2 times 1 equals 24.

Exam Tip: When forming words using all letters of a word with no repetition, calculate the arrangement as (total letters)!.

Question 9. For a group photograph, 3 boys and 2 girls stand in a line in all possible ways. How many photos could be taken if each photo corresponds to each such arrangement?
Answer: There are a total of 5 people (3 boys and 2 girls) to arrange in a line. The first position can be filled by any of the 5 people in 5 ways. The second position can be filled by any of the remaining 4 people in 4 ways. The third position can be filled in 3 ways, the fourth in 2 ways, and the fifth in 1 way. By the multiplication principle of counting, the total number of arrangements is 5 × 4 × 3 × 2 × 1 = 120. Therefore, 120 photos could be taken.
In simple words: Arrange 5 people in a line: first spot gets any of 5 (5 ways), second gets any of 4 (4 ways), third gets any of 3 (3 ways), fourth gets any of 2 (2 ways), fifth gets the last one (1 way). Multiply: 5 times 4 times 3 times 2 times 1 equals 120.

Exam Tip: The number of ways to arrange n distinct objects in a line is n!. The genders or other characteristics of the objects do not change this count.

Question 10. Six pictures are to be arranged (in line from left to right) on a wall of an art gallery for display. How many arrangements are possible?
Answer: There are 6 distinct pictures to arrange in a line. The first position can be filled by any of the 6 pictures in 6 ways. The second position can be filled by any of the remaining 5 pictures in 5 ways. Continuing this pattern, the total number of arrangements is 6 × 5 × 4 × 3 × 2 × 1 = 720.
In simple words: Arrange 6 pictures in order. First position: 6 choices. Second position: 5 choices. Keep going: 4, 3, 2, 1. Multiply all: 6 times 5 times 4 times 3 times 2 times 1 equals 720.

Exam Tip: Arranging n distinct objects in a line always gives n! arrangements.

Question 11. Sandy has 5 shirts, 4 pants, 3 pairs of socks and 2 pairs of shoes. In how many different ways can he dress himself?
Answer: Sandy can choose a shirt in 5 ways, a pair of pants in 4 ways, a pair of socks in 3 ways, and a pair of shoes in 2 ways. Since these choices are independent and must all be made together, by the multiplication principle of counting, the total number of ways he can dress himself is 5 × 4 × 3 × 2 = 120.
In simple words: Pick a shirt (5 ways), pick pants (4 ways), pick socks (3 ways), pick shoes (2 ways). Multiply: 5 times 4 times 3 times 2 equals 120.

Exam Tip: When dressing or assembling outfits by choosing one item from each category, multiply the number of choices in each category.

Question 12. How many 3-letter code words are possible using the first 10 letters of English alphabet if
(i) no letter can be repeated?
(ii) letters can be repeated?
Answer:
(i) No letter can be repeated.
The first position can be filled in 10 ways. The second position can be filled in 9 ways (since one letter is already used). The third position can be filled in 8 ways. By the multiplication principle of counting, the total number of 3-letter code words is 10 × 9 × 8 = 720.

(ii) Letters can be repeated.
The first position can be filled in 10 ways. The second position can be filled in 10 ways (since letters can repeat). The third position can be filled in 10 ways. By the multiplication principle of counting, the total number of 3-letter code words is 10 × 10 × 10 = 1000.
In simple words: Without repeats: first spot gets 10 choices, second gets 9, third gets 8 - multiply to get 720. With repeats: each spot gets 10 choices - multiply 10 three times to get 1000.

Exam Tip: Without repetition, choices decrease at each position. With repetition allowed, choices remain the same at each position.

Question 13. How many different 5-letter code words can be formed using the first 10 letters of the English alphabet, if no letter can be repeated?
Answer: The first position can be filled in 10 ways by any of the 10 letters. The second position can be filled in 9 ways by any of the remaining 9 letters. The third position can be filled in 8 ways. The fourth position can be filled in 7 ways. The fifth position can be filled in 6 ways. By the multiplication principle of counting, the total number of 5-letter code words is 10 × 9 × 8 × 7 × 6 = 30240.
In simple words: Choose letters without repeating. First position: 10 choices. Second: 9. Third: 8. Fourth: 7. Fifth: 6. Multiply all: 10 times 9 times 8 times 7 times 6 equals 30240.

Exam Tip: When choosing r items from n distinct items without repetition and order matters, the count is n × (n-1) × (n-2) × ... × (n-r+1).

Question 14. How many outcomes are possible when a coin is tossed
(i) 2 times?
(ii) 3 times?
(iii) 4 times?
(iv) 5 times?
Answer:
(i) When a coin is tossed 2 times, each toss has 2 possible outcomes (heads or tails). By the multiplication principle of counting, the total number of outcomes is 2 × 2 = 4.

(ii) When a coin is tossed 3 times, the total number of outcomes is 2 × 2 × 2 = 8.

(iii) When a coin is tossed 4 times, the total number of outcomes is 2 × 2 × 2 × 2 = 16.

(iv) When a coin is tossed 5 times, the total number of outcomes is 2 × 2 × 2 × 2 × 2 = 32.
In simple words: Each toss has 2 outcomes (heads or tails). For 2 tosses: 2 times 2 equals 4. For 3 tosses: 2^3 equals 8. For 4 tosses: 2^4 equals 16. For 5 tosses: 2^5 equals 32.

Exam Tip: When an event with k outcomes is repeated n times, the total number of outcome sequences is k^n.

Question 15. Given 6 flags of different colours. How many different signals can be generated by hoisting the flags on a vertical pole (one below the other) if each signal requires the use of
(i) 2 flags?
(ii) 3 flags?
(iii) 4 flags?
Answer:
(i) The first flag (upper position) can be chosen in 6 ways. The second flag (lower position) can be chosen in 5 ways. By the multiplication principle of counting, the number of signals using 2 flags is 6 × 5 = 30.

(ii) The first position can be filled in 6 ways, the second in 5 ways, and the third in 4 ways. By the multiplication principle of counting, the number of signals using 3 flags is 6 × 5 × 4 = 120.

(iii) The first position can be filled in 6 ways, the second in 5 ways, the third in 4 ways, and the fourth in 3 ways. By the multiplication principle of counting, the number of signals using 4 flags is 6 × 5 × 4 × 3 = 360.
In simple words: For 2 flags: pick one of 6 for top (6 ways), one of 5 remaining for below (5 ways) - multiply to get 30. For 3 flags: 6 times 5 times 4 equals 120. For 4 flags: 6 times 5 times 4 times 3 equals 360.

Exam Tip: When arranging r items from n distinct items in a line (order matters), the count is n × (n-1) × ... × (n-r+1).

Question 16. How many 3-digit numbers can be formed from the digits 1, 2, 3, 4 and 5 if
(i) repetition of digits is allowed?
(ii) repetition of digits is not allowed?
Answer:
(i) Repetition of digits is allowed.
The first digit (hundreds place) can be any of the 5 digits in 5 ways. The second digit (tens place) can be any of the 5 digits in 5 ways. The third digit (units place) can be any of the 5 digits in 5 ways. By the multiplication principle of counting, the total number of 3-digit numbers is 5 × 5 × 5 = 125.

(ii) Repetition of digits is not allowed.
The first digit (hundreds place) can be any of the 5 digits in 5 ways. The second digit (tens place) can be any of the remaining 4 digits in 4 ways. The third digit (units place) can be any of the remaining 3 digits in 3 ways. By the multiplication principle of counting, the total number of 3-digit numbers is 5 × 4 × 3 = 60.
In simple words: With repeats: each place can be any of 5 digits - multiply 5 three times to get 125. Without repeats: first place gets 5 choices, second gets 4, third gets 3 - multiply to get 60.

Exam Tip: With repetition allowed, use the same count for each position. Without repetition, reduce the available count at each step.

Question 17. How many four digit numbers can be formed in which all the digits are different?
Answer: The digits available are 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. Since a 4-digit number cannot start with 0, the first digit (thousands place) can be any of the 9 digits (1-9) in 9 ways. The second digit (hundreds place) can be any of the remaining 9 digits (including 0 if not used, but excluding the first digit) in 9 ways. The third digit (tens place) can be any of the remaining 8 digits in 8 ways. The fourth digit (units place) can be any of the remaining 7 digits in 7 ways. By the multiplication principle of counting, the total number of 4-digit numbers with all different digits is 9 × 9 × 8 × 7 = 4536.
In simple words: The first digit is 1-9 (not 0), giving 9 ways. The second digit is any remaining digit including 0 (9 ways). Third digit: 8 remaining. Fourth digit: 7 remaining. Multiply: 9 times 9 times 8 times 7 equals 4536.

Exam Tip: When forming numbers without repetition, handle the leading zero restriction first, then count decreasing choices for remaining positions.

Question 18. How many three digit numbers can be formed without using the digits 2, 3, 5, 6, 7 and 9?
Answer: The available digits are 0, 1, 4, 8 - a total of 4 digits. For a 3-digit number, the first digit (hundreds place) cannot be 0, so it can be any of 1, 4, 8 in 3 ways. The second digit (tens place) can be any of the 4 available digits in 4 ways. The third digit (units place) can be any of the 4 available digits in 4 ways. By the multiplication principle of counting, the total number of 3-digit numbers is 3 × 4 × 4 = 48.
In simple words: Available digits are 0, 1, 4, 8. First digit (not 0): 1, 4, or 8 - 3 choices. Second digit: any of 0, 1, 4, 8 - 4 choices. Third digit: any of 0, 1, 4, 8 - 4 choices. Multiply: 3 times 4 times 4 equals 48.

Exam Tip: First identify which digits are available after the restrictions. Then count arrangements with the non-zero constraint for the leading digit.

Question 19. How many three digit numbers can be formed using the digits 0, 1, 3, 5, 6, 7 if
(i) repetition of digits is not allowed?
(ii) repetition of digits is allowed?
Answer:
(i) Repetition of digits is not allowed.
There are 6 available digits. The first digit (hundreds place) cannot be 0, so it can be any of 1, 3, 5, 6, 7 in 5 ways. The second digit (tens place) can be any of the remaining 5 digits (including 0 if not used) in 5 ways. The third digit (units place) can be any of the remaining 4 digits in 4 ways. By the multiplication principle of counting, the total number of 3-digit numbers is 5 × 5 × 4 = 100.

(ii) Repetition of digits is allowed.
The first digit (hundreds place) cannot be 0, so it can be any of 1, 3, 5, 6, 7 in 5 ways. The second digit (tens place) can be any of the 6 digits in 6 ways. The third digit (units place) can be any of the 6 digits in 6 ways. By the multiplication principle of counting, the total number of 3-digit numbers is 5 × 6 × 6 = 180.
In simple words: Without repeats: first digit is 1, 3, 5, 6, or 7 (5 ways), second is any of 5 remaining, third is any of 4 remaining - multiply to get 100. With repeats: first is 1, 3, 5, 6, or 7 (5 ways), second is any of the 6 (6 ways), third is any of the 6 (6 ways) - multiply to get 180.

Exam Tip: Always handle the non-zero leading digit first, then apply the repetition rule to the remaining positions.

Question 20. How many 2-digit even numbers can be formed from the digits 1, 2, 3, 4 and 5 if the digits
(i) can repeat?
(ii) cannot repeat?
Answer:
(i) Digits can repeat.
For the number to be even, the units digit must be 2 or 4. So the units place can be filled in 2 ways. The tens place can be any of the 5 digits in 5 ways. By the multiplication principle of counting, the total number of 2-digit even numbers is 5 × 2 = 10.

(ii) Digits cannot repeat.
For the number to be even, the units digit must be 2 or 4. So the units place can be filled in 2 ways. The tens place can be any of the remaining 4 digits in 4 ways. By the multiplication principle of counting, the total number of 2-digit even numbers is 4 × 2 = 8.
In simple words: With repeats: units must be 2 or 4 (2 ways), tens can be any digit (5 ways) - multiply to get 10. Without repeats: units must be 2 or 4 (2 ways), tens can be any of 4 remaining (4 ways) - multiply to get 8.

Exam Tip: For even numbers, fill the units place first with even digits only. For odd numbers, fill the units place first with odd digits only. Then fill remaining places.

7.2 Factorial Notation

Factorial. The continued product of first n natural numbers is called n factorial or factorial n and is denoted by \( \underline{n} \) or \( n! \)

Thus, \( \underline{n} \) or \( n! = 1 \times 2 \times 3 \times 4 \times \ldots \times (n-1)n = n(n-1)(n-2) \times \ldots \times 3 \times 2 \times 1 \) (in reverse order)

For example:

\( \underline{1} = 1 \)
\( \underline{2} = 1 \times 2 = 2 \)
\( \underline{3} = 1 \times 2 \times 3 = 6 \)
\( \underline{4} = 1 \times 2 \times 3 \times 4 = 24 \)
\( \underline{5} = 1 \times 2 \times 3 \times 4 \times 5 = 120 \) and so on.

Note that \( \underline{n} = n(n-1)(n-2) \times \ldots \times 3 \times 2 \times 1 = n \underline{n-1} \) for n > 1

\( = n(n-1) \underline{n-2} \) for n > 2

\( = n(n-1)(n-2) \underline{n-3} \) for n > 3 and so on.

For example, \( \underline{8} = 8 \times \underline{7} = 8 \times 7 \times \underline{6} \) and so on.

Meaning of Zero Factorial. According to the above definition, \( \underline{0} \) makes no sense. However, we define \( \underline{0} = 1 \). It will be clear from the subsequent work why we make this convention.

Note: When n is a negative integer or a fraction, n factorial is not defined.

Illustrative Examples

Example 1. Find the value of
(i) \( \frac{\underline{25}}{\underline{23}} \)
(ii) \( \frac{\underline{9}}{\underline{6} \times \underline{3}} \)
(iii) (7 - 4)!
(iv) \( \underline{7} - \underline{5} \)
(v) \( \underline{2} + \underline{3} \)
Answer:

(i) \( \frac{\underline{25}}{\underline{23}} = \frac{25 \times 24 \times \underline{23}}{\underline{23}} = 25 \times 24 = 600 \)

(ii) \( \frac{\underline{9}}{\underline{6} \times \underline{3}} = \frac{9 \times 8 \times 7 \times \underline{6}}{\underline{6} \times \underline{3}} = \frac{9 \times 8 \times 7}{\underline{3}} = \frac{9 \times 8 \times 7}{1 \times 2 \times 3} = 84 \)

(iii) (7 - 4)! = 3! = 1 × 2 × 3 = 6

(iv) \( \underline{7} - \underline{5} = 1 \times 2 \times 3 \times 4 \times 5 \times 6 \times 7 - 1 \times 2 \times 3 \times 4 \times 5 = 5040 - 120 = 4920 \)

(v) \( \underline{2} + \underline{3} = 1 \times 2 + 1 \times 2 \times 3 = 2 + 6 = 8 \)
In simple words: For part (i), cancel \( \underline{23} \) and multiply 25 times 24. For part (ii), cancel \( \underline{6} \) and divide the remaining product by \( \underline{3} \). For parts (iv) and (v), compute each factorial separately, then subtract or add.

Exam Tip: When dividing factorials, look for common terms to cancel. When adding or subtracting factorials, compute each separately first.

Example 2. Evaluate \( \frac{n!}{(n-r)!} \) when
(i) n = 9, r = 5
(ii) r = 3
Answer:

(i) When n = 9, r = 5:

\( \frac{n!}{(n-r)!} = \frac{\underline{9}}{(9-5)!} = \frac{\underline{9}}{\underline{4}} = \frac{9 \times 8 \times 7 \times 6 \times 5 \times \underline{4}}{\underline{4}} = 9 \times 8 \times 7 \times 6 \times 5 = 15120 \)

(ii) When r = 3:

\( \frac{n!}{(n-r)!} = \frac{n!}{(n-3)!} = \frac{n(n-1)(n-2)(n-3)!}{(n-3)!} = n(n-1)(n-2) \)
In simple words: For part (i), expand both factorials and cancel the common \( \underline{4} \) term. For part (ii), write out the pattern showing how \( \underline{n} \) expanded through (n-3)! cancels.

Exam Tip: The fraction \( \frac{n!}{(n-r)!} \) always simplifies to a product of r consecutive integers starting from n going downward.

Example 3. Evaluate \( \frac{\underline{n}}{r! \underline{n-r}} \) when
(i) n = 7, r = 5
(ii) n = 52, r = 48
(iii) r = 2
Answer:

(i) When n = 7, r = 5:

\( \frac{\underline{n}}{r! \underline{n-r}} = \frac{\underline{7}}{\underline{5} \underline{2}} = \frac{7 \times 6 \times \underline{5}}{\underline{5} \times 2} = \frac{7 \times 6}{2} = \frac{42}{2} = 21 \)

(ii) When n = 52, r = 48:

\( \frac{\underline{n}}{r! \underline{n-r}} = \frac{\underline{52}}{\underline{48} \underline{4}} = \frac{52 \times 51 \times 50 \times 49 \times \underline{48}}{\underline{48} \times 4!} = \frac{52 \times 51 \times 50 \times 49}{1 \times 2 \times 3 \times 4} = 270725 \)

(iii) When r = 2:

\( \frac{\underline{n}}{r! \underline{n-r}} = \frac{\underline{n}}{\underline{2} \underline{n-2}} = \frac{n(n-1)(n-2)!}{\underline{2} (n-2)!} = \frac{n(n-1)}{1 \times 2} = \frac{n(n-1)}{2} \)
In simple words: For part (i), expand and cancel \( \underline{5} \) from numerator and denominator. For part (ii), expand the numerator product and divide by 4!. For part (iii), show the general pattern for r = 2.

Exam Tip: The expression \( \frac{n!}{r!(n-r)!} \) is the binomial coefficient, often written as C(n,r) or \( \binom{n}{r} \).

Example 4. Convert the following products into factorials:
(i) 5 × 6 × 7 × 8 × 9
(ii) 2 × 4 × 6 × 8 × 10 × 12
Answer:

(i) \( 5 \times 6 \times 7 \times 8 \times 9 = \frac{1 \times 2 \times 3 \times 4 \times 5 \times 6 \times 7 \times 8 \times 9}{1 \times 2 \times 3 \times 4} = \frac{\underline{9}}{\underline{4}} \)

(ii) \( 2 \times 4 \times 6 \times 8 \times 10 \times 12 = (2 \times 1)(2 \times 2)(2 \times 3)(2 \times 4)(2 \times 5)(2 \times 6) = 2^6 \times (1 \times 2 \times 3 \times 4 \times 5 \times 6) = 2^6 \times \underline{6} \)
In simple words: For part (i), recognize this is 9! divided by 4! because 9! contains all terms 1 through 9, and you need only 5 through 9. For part (ii), factor out 2 from each term to get 2^6 times 6!.

Exam Tip: To convert a product to factorial form, identify what factors are missing from a complete factorial expansion, then divide by the factorial of those missing factors.

Example 5. Find n, if
(i) \( \frac{1}{\underline{8}} + \frac{1}{\underline{9}} = \frac{n}{\underline{10}} \)
(ii) (n + 1)! = 12 × (n - 1)!
Answer:

(i) \( \frac{1}{\underline{8}} + \frac{1}{\underline{9}} = \frac{n}{\underline{10}} \)

\( \frac{1}{\underline{8}} + \frac{1}{\underline{9}} = \frac{n}{\underline{10}} \Rightarrow \frac{1}{\underline{8}} + \frac{1}{9 \times \underline{8}} = \frac{n}{10 \times 9 \times \underline{8}} \)

\( \Rightarrow \frac{1}{\underline{8}}\left(1 + \frac{1}{9}\right) = \frac{n}{10 \times 9 \times \underline{8}} \Rightarrow 1 + \frac{1}{9} = \frac{n}{90} \)

\( \Rightarrow \frac{10}{9} = \frac{n}{90} \Rightarrow n = 100 \)

(ii) (n + 1)! = 12 × (n - 1)!

\( \Rightarrow (n + 1) \times n \times (n - 1)! = 12 \times (n - 1)! \)

\( \Rightarrow (n + 1) \times n = 12 \Rightarrow n^2 + n - 12 = 0 \)

\( \Rightarrow (n - 3)(n + 4) = 0 \Rightarrow n - 3 = 0 \text{ or } n + 4 = 0 \)

\( \Rightarrow n = 3 \text{ or } n = -4 \)

But n cannot be negative as n! is not meaningful when n is negative. Therefore, n = 3.
In simple words: For part (i), rewrite each factorial in terms of the others by cancellation. For part (ii), expand (n+1)! as (n+1) × n × (n-1)!, then cancel (n-1)! and solve the resulting quadratic equation.

Exam Tip: When solving factorial equations, try to factor out and cancel common factorial terms. Then solve the simplified algebraic equation.

Example 6. Find n, if \( \frac{\underline{n}}{2 \underline{n-2}} \) and \( \frac{\underline{n}}{4 \underline{n-4}} \) are in the ratio 2 : 1.
Answer:

Given: \( \frac{\underline{n}}{2 \underline{n-2}} : \frac{\underline{n}}{4 \underline{n-4}} = 2 : 1 \)

\( \Rightarrow \frac{\underline{n}}{2 \underline{n-2}} \times \frac{4 \underline{n-4}}{\underline{n}} = \frac{2}{1} \)

\( \Rightarrow \frac{4 \times 3 \times \underline{2} \times \underline{n-4}}{2 \underline{n-2} \times (n-4) \underline{n-4}} = \frac{2}{1} \)

\( \Rightarrow \frac{4 \times 3}{2(n-2)(n-3)} = \frac{2}{1} \)

\( \Rightarrow \frac{12}{2(n-2)(n-3)} = \frac{2}{1} \Rightarrow (n-2)(n-3) = 6 \)

\( \Rightarrow n^2 - 5n = 0 \Rightarrow n(n-5) = 0 \)

\( \Rightarrow n = 0 \text{ or } n = 5 \)

But, for n = 0, \( \underline{n-2} \) and \( \underline{n-4} \) are not meaningful. Therefore, n = 5.
In simple words: Set up the ratio as an equation. Cross-multiply and expand the factorials. Cancel common terms to simplify the equation, then solve for n.

Exam Tip: When two factorial expressions are in a given ratio, set up an equation and cross-multiply. Expand carefully and look for opportunities to cancel.

Example 7. Prove that \( \underline{2n} = 1.3.5. \ldots (2n-1).2^n.\underline{n} \)
Answer:

\( \underline{2n} = 1.2.3.4.5.6. \ldots (2n-1)(2n) \)

\( = [1.3.5. \ldots (2n-1)][2.4.6. \ldots 2n] \)

\( = [1.3.5. \ldots (2n-1)][(2.1)(2.2)(2.3) \ldots (2.n)] \)

\( = 1.3.5. \ldots (2n-1).2^n.(1.2.3 \ldots n) \)

\( = 1.3.5. \ldots (2n-1).2^n.\underline{n} \), as desired.
In simple words: Split 2n! into odd factors and even factors separately. The even factors can be written as 2 times each number from 1 to n, giving 2^n times n!. Combine to get the desired result.

Exam Tip: When proving factorial identities, separate the factors by type (odd/even, specified range/remainder) and regroup them to match the target form.

Exercise 7.2

Very Short Answer Type Questions (1 to 5):

Question 1. Find the values of
(i) \( \underline{6} \)
(ii) \( \underline{4} + \underline{3} \)
(iii) \( \underline{9} - \underline{6} \)
(iv) \( \underline{6} - \underline{4} \)
(v) \( \underline{3} \times \underline{5} \)
(vi) \( \frac{\underline{8}}{8 \underline{6}} \)
(vii) \( \frac{\underline{12}}{\underline{10} \underline{2}} \)
(viii) \( \frac{\underline{9} - \underline{8}}{\underline{7}} \)
(ix) \( \frac{\underline{n+2}}{\underline{n+1}} \)
Answer:

(i) \( \underline{6} = 1 \times 2 \times 3 \times 4 \times 5 \times 6 = 720 \)

(ii) \( \underline{4} + \underline{3} = (1 \times 2 \times 3 \times 4) + (1 \times 2 \times 3) = 24 + 6 = 30 \)

(iii) \( \underline{9} - \underline{6} = 362880 - 720 = 362160 \)

(iv) \( \underline{6} - \underline{4} = 720 - 24 = 696 \)

(v) \( \underline{3} \times \underline{5} = 6 \times 120 = 720 \)

(vi) \( \frac{\underline{8}}{8 \underline{6}} = \frac{8 \times \underline{7} \times \underline{6}}{8 \times \underline{6}} = 7 \)

(vii) \( \frac{\underline{12}}{\underline{10} \underline{2}} = \frac{12 \times 11 \times \underline{10}}{\underline{10} \times 2 \times 1} = \frac{132}{2} = 66 \)

(viii) \( \frac{\underline{9} - \underline{8}}{\underline{7}} = \frac{(9 \times \underline{8}) - \underline{8}}{\underline{7}} = \frac{\underline{8}(9 - 1)}{\underline{7}} = \frac{40320 \times 8}{5040} = 64 \)

(ix) \( \frac{\underline{n+2}}{\underline{n+1}} = \frac{(n+2) \times \underline{n+1}}{\underline{n+1}} = n + 2 \)
In simple words: For (i), multiply 1 × 2 × 3 × 4 × 5 × 6. For (ii), find 4! and 3! separately, then add. For the fractions, expand and cancel common factorial terms.

Exam Tip: Always compute each factorial from first principles if needed. Look for opportunities to cancel when dividing factorials.

Question 6. Convert into factorials:
(i) 3 × 6 × 9 × 12 × 15 × 18
(ii) 6 × 7 × 8 × 9 × 10 × 11 × 12
Answer:
(i) 3 × 6 × 9 × 12 × 15 × 18 = (3 × 1) × (3 × 2) × (3 × 3) × (3 × 4) × (3 × 5) × (3 × 6)

\[ = 3^6 (1 × 2 × 3 × 4 × 5 × 6) = 3^6 \times 6! \]

(ii) 6 × 7 × 8 × 9 × 10 × 11 × 12 = \[ \frac{12!}{5!} \]
In simple words: Factor out common terms and express products as ratios of factorials or powers times factorials.

Exam Tip: Look for patterns - products with constant differences between terms can often be written as factorials divided by smaller factorials, or as a power multiplied by a factorial.

Question 7. Find the value of n, if
(i) \( \binom{n+1}{5} = \binom{n-1}{4} \)
(ii) \( \frac{1}{\binom{5}{6}} + \frac{1}{\binom{6}{n}} = \frac{n}{16} \)
(iii) \( \frac{1}{\binom{6}{7}} + \frac{1}{\binom{8}{n}} = \frac{n}{18} \)
(iv) \( \binom{n}{n-2} = 110 \)
(v) \( \binom{n}{2} \) and \( \binom{2n}{4} \) are in the ratio 1 : 6
Answer:
(i) This equation is impossible since \( \binom{n+1}{5} \) and \( \binom{n-1}{4} \) have no valid solution (binomial coefficients require positive arguments within range).

(ii) Both terms on the left are zero or undefined for the given indices. This part requires clarification from the original text, as standard binomial notation does not support these indices.

(iii) Similarly, this part involves non-standard binomial indices and cannot be solved as stated.

(iv) \( \binom{n}{n-2} = \binom{n}{2} = \frac{n(n-1)}{2} = 110 \)

\[ n(n-1) = 220 \implies n^2 - n - 220 = 0 \implies (n-15)(n+14) = 0 \]

\[ \therefore n = 15 \text{ (since } n \text{ must be positive)} \]

(v) \( \binom{n}{2} : \binom{2n}{4} = 1 : 6 \)

\[ \frac{\binom{n}{2}}{\binom{2n}{4}} = \frac{1}{6} \]

\[ \frac{\frac{n(n-1)}{2}}{\frac{2n(2n-1)(2n-2)(2n-3)}{24}} = \frac{1}{6} \]

\[ \frac{n(n-1)}{2} \times \frac{24}{2n(2n-1)(2n-2)(2n-3)} = \frac{1}{6} \]

\[ \frac{12n(n-1)}{2n(2n-1)(2n-2)(2n-3)} = \frac{1}{6} \]

\[ 72n(n-1) = 2n(2n-1)(2n-2)(2n-3) \]

\[ 36(n-1) = (2n-1)(2n-2)(2n-3) \]

\[ 36(n-1) = 2(2n-1)(n-1)(2n-3) \]

\[ 18 = (2n-1)(2n-3) \implies 4n^2 - 8n + 3 = 18 \]

\[ 4n^2 - 8n - 15 = 0 \implies (2n-5)(2n+3) = 0 \]

\[ n = \frac{5}{2} \text{ or } n = -\frac{3}{2} \]

Since n must be a positive integer, there is no valid solution for part (v) as stated, or the problem requires re-examination.
In simple words: Use the formulas for binomial coefficients and factorial notation to set up equations, then solve for n by simplifying and factoring.

Exam Tip: Always check that the value of n satisfies the constraint that binomial coefficient indices are valid (non-negative integers with the lower index ≤ upper index). Reject solutions that violate these conditions.

7.3 PERMUTATIONS

A permutation is any arrangement that can be created by selecting some or all items from a collection of distinct objects.

For example:

(i) All arrangements possible from the letters a, b, c when selecting 2 at a time are: ab, ba, bc, cb, ca, ac. Thus, there are 6 permutations of 3 distinct objects taken 2 at a time.

(ii) All arrangements possible from the letters C, A, T when selecting all three are: CAT, CTA, ATC, ACT, TAC, TCA. Thus, there are 6 permutations of 3 distinct objects taken all at a time.

Important note: In permutations, the sequence in which objects are arranged matters. Changing the order produces a different permutation. In example (i), ab and ba are distinct, and in (ii), CAT, CTA, and ATC are all different permutations.

If r and n are two positive whole numbers with 1 ≤ r ≤ n, the number of ways to arrange n distinct objects taken r at a time (without repetition) is denoted as nPr or P(n, r).

7.3.1 Permutations when all objects are different

Theorem 1. The number of ways to arrange n distinct objects taken r at a time without repetition is given by:

\[ P(n, r) = n(n - 1)(n - 2) \ldots (n - r + 1) = \frac{n!}{(n - r)!}, \quad 1 \leq r \leq n \]

Proof. The number of permutations of n distinct objects taken r at a time equals the number of methods to fill r empty slots using n distinct objects.

The first slot can be filled in n ways (any of the n objects can go there). Once the first slot is filled, (n - 1) objects remain. Any one of them fits in the second slot. The third slot can be filled using n - 2 objects, the fourth with n - 3 objects, and so on. The rth slot can be filled in n - (r - 1) = n - r + 1 ways. Using the fundamental counting principle, the r slots can be filled in n(n - 1)(n - 2) ... (n - r + 1) ways.

\[ \therefore \text{ } ^n P_r = n(n - 1)(n - 2) \ldots (n - r + 1) \]

\[ = \frac{n(n - 1)(n - 2) \ldots (n - r + 1)(n - r) \ldots 3 \times 2 \times 1}{(n - r)(n - r - 1) \ldots 3 \times 2 \times 1} \]

(Multiplying both numerator and denominator by (n - r)(n - r - 1) ... 3 × 2 × 1)

\[ = \frac{n!}{(n - r)!}, \quad 1 \leq r \leq n \]

Corollary 1. \( ^n P_n = n(n - 1) \ldots (n - n + 1) = n(n - 1) \ldots 1 = n! \)

Corollary 2. Substituting r = n into \( ^n P_r = \frac{n!}{(n - r)!} \), we get \( \frac{n!}{0!} = n! \). This justifies defining \( 0! = 1 \).

REMARK: We define \( ^n P_0 = 1 \) - that is, the number of arrangements of n distinct objects taking nothing is set as 1. Counting permutations is simply counting the ways some or all objects can be arranged. Arranging zero objects is the same as choosing to leave all objects behind, and there is exactly one way to do this. That is why \( ^n P_0 = 1 \). The formula \( ^n P_r = \frac{n!}{(n - r)!} \) works for r = 0 as well.

Theorem 2. The number of permutations of n distinct objects taken r at a time with repetition allowed is \( n^r \).

Proof. The number of permutations of n distinct objects taken r at a time with repetition equals the number of ways to fill r empty slots from n distinct objects. The first slot can be filled with any one of the n objects. Once filled, the second slot can also be filled in n ways since we are free to use the same object again. When the first two slots are filled in n × n ways, the third slot can be filled in n ways as well, and similarly for the rest. By the fundamental counting principle, r slots can be filled in (n × n × ... r times) = nr ways.

Therefore, the required number of permutations = \( n^r \).

ILLUSTRATIVE EXAMPLES

Question. Example 1. Find the values of
(i) \( ^5 P_5 \)
(ii) \( ^5 P_3 \)
(iii) \( ^{60} P_{48} \)
(iv) \( ^{60} P_2 \)
Answer:
(i) \( ^5 P_5 = 5 \times 4 \times 3 \times 2 \times 1 = 120 \)

(ii) \( ^5 P_3 = 5 \times 4 \times 3 = 60 \)

(iii) \( ^{60} P_{48} = \frac{60!}{(60 - 48)!} = \frac{60!}{12!} \)

(iv) \( ^{60} P_2 = 60 \times 59 = 3540 \)
In simple words: Apply the permutation formula - multiply descending integers starting from n for r terms, or use the factorial formula \( \frac{n!}{(n-r)!} \).

Exam Tip: For small values of r, it is faster to multiply directly rather than computing full factorials. Always verify your arithmetic, especially in products of many terms.

Question. Example 2. Find n if
(i) \( ^n P_4 : ^n P_5 = 1 : 2 \)
(ii) \( ^n P_4 : ^{n-1} P_3 = 9 : 1 \)
(iii) \( P(n, 4) = 20 P(n, 2) \)
Answer:
(i) Given \( ^n P_4 : ^n P_5 = 1 : 2 \)

\[ \frac{^n P_4}{^n P_5} = \frac{1}{2} \implies \frac{n(n-1)(n-2)(n-3)}{n(n-1)(n-2)(n-3)(n-4)} = \frac{1}{2} \]

\[ \frac{1}{n-4} = \frac{1}{2} \implies n - 4 = 2 \implies n = 6 \]

(ii) Given \( ^n P_4 : ^{n-1} P_3 = 9 : 1 \)

\[ \frac{^n P_4}{^{n-1} P_3} = \frac{9}{1} \implies \frac{n(n-1)(n-2)(n-3)}{(n-1)(n-2)(n-3)} = 9 \]

\[ n = 9 \]

(iii) Given \( P(n, 4) = 20 P(n, 2) \)

\[ n(n-1)(n-2)(n-3) = 20n(n-1) \]

\[ (n-2)(n-3) = 20 \implies n^2 - 5n - 14 = 0 \]

\[ (n+2)(n-7) = 0 \implies n = -2 \text{ or } n = 7 \]

Since n must be positive, \( n = 7 \).
In simple words: Write the permutation expressions using the formula, simplify by canceling common factors, then solve for n algebraically.

Exam Tip: When ratios of permutations are given, many terms cancel out. Identify and remove these cancellations first to simplify your equation before solving.

Question. Example 3. (i) Find n if P(n, 4) = 2P(5, 3)
(ii) Find r if P(10, r) = 720
Answer:
(i) Given \( P(n, 4) = 2P(5, 3) \)

\[ n(n-1)(n-2)(n-3) = 2 \times 5 \times 4 \times 3 \]

\[ n(n-1)(n-2)(n-3) = 5 \times 4 \times 3 \times 2 \]

\[ n = 5 \]

(ii) Given \( P(10, r) = 720 \)

\[ P(10, r) = 10 \times 9 \times 8 = 720 \]

\[ r = 3 \]
In simple words: Evaluate the known permutation and then match it to see which product of descending integers gives that result.

Exam Tip: Recognize patterns - 720 = 10 × 9 × 8, so r = 3. Breaking down target numbers into their prime factors can help identify the right value of r quickly.

Question. Example 4. Find n if:
(i) \( ^n P_5 = 42 \times ^n P_3 \)
(ii) \( ^{2n+1} P_{n-1} : ^{2n-1} P_n = 3 : 5 \)
Answer:
(i) Given \( ^n P_5 = 42 \times ^n P_3 \)

\[ n(n-1)(n-2)(n-3)(n-4) = 42 \times n(n-1)(n-2) \]

\[ (n-3)(n-4) = 42 \]

\[ n^2 - 7n + 12 = 42 \]

\[ n^2 - 7n - 30 = 0 \]

\[ (n-10)(n+3) = 0 \]

\[ n = 10 \text{ or } n = -3 \]

Since n must be positive, \( n = 10 \).

(ii) Given \( ^{2n+1} P_{n-1} : ^{2n-1} P_n = 3 : 5 \)

\[ \frac{^{2n+1} P_{n-1}}{^{2n-1} P_n} = \frac{3}{5} \]

\[ \frac{\frac{(2n+1)!}{(2n+1-(n-1))!}}{\frac{(2n-1)!}{(2n-1-n)!}} = \frac{3}{5} \]

\[ \frac{\frac{(2n+1)!}{(n+2)!}}{\frac{(2n-1)!}{(n-1)!}} = \frac{3}{5} \]

\[ \frac{(2n+1)!}{(n+2)!} \times \frac{(n-1)!}{(2n-1)!} = \frac{3}{5} \]

\[ \frac{(2n+1) \times 2n \times (2n-1)!}{(n+2)(n+1)n(n-1)!} \times \frac{(n-1)!}{(2n-1)!} = \frac{3}{5} \]

\[ \frac{(2n+1) \times 2n}{(n+2)(n+1)n} = \frac{3}{5} \]

\[ \frac{2(2n+1)}{(n+2)(n+1)} = \frac{3}{5} \]

\[ 10(2n+1) = 3(n+2)(n+1) \]

\[ 20n + 10 = 3(n^2 + 3n + 2) \]

\[ 20n + 10 = 3n^2 + 9n + 6 \]

\[ 3n^2 - 11n - 4 = 0 \]

\[ (n-4)(3n+1) = 0 \]

\[ n = 4 \text{ or } n = -\frac{1}{3} \]

Since n must be a positive integer, \( n = 4 \).
In simple words: Cancel common factors from the expanded permutation expressions, then set up an equation based on the given ratio and solve for n.

Exam Tip: Expand permutation factorials carefully and cancel identical terms before setting up the ratio equation. This reduces calculation errors significantly.

Question. Example 5. Find the value of r if
(i) \( 5 \times ^4 P_r = 6 \times ^5 P_{r-1} \)
(ii) \( P(10, r+1) : P(11, r) = 30 : 11 \)
Answer:
(i) Given \( 5 \times ^4 P_r = 6 \times ^5 P_{r-1} \)

\[ 5 \times \frac{4!}{(4-r)!} = 6 \times \frac{5!}{(5-(r-1))!} \]

\[ 5 \times \frac{4!}{(4-r)!} = 6 \times \frac{5!}{(6-r)!} \]

\[ \frac{5!}{(4-r)!} = \frac{6 \times 5!}{(6-r)!} \]

(Since \( 5 \times 4! = 5! \))

\[ \frac{1}{(4-r)!} = \frac{6}{(6-r)!} \]

\[ (6-r)! = 6(4-r)! \]

\[ (6-r)(5-r) = 6 \]

\[ 30 - 11r + r^2 = 6 \]

\[ r^2 - 11r + 24 = 0 \]

\[ (r-3)(r-8) = 0 \]

\[ r = 3 \text{ or } r = 8 \]

Since \( ^4 P_r \) and \( ^5 P_{r-1} \) must be defined, we need \( r \leq 4 \) and \( r - 1 \leq 5 \), so \( r \leq 4 \). Rejecting \( r = 8 \), the answer is \( r = 3 \).

(ii) Given \( P(10, r+1) : P(11, r) = 30 : 11 \)

\[ \frac{P(10, r+1)}{P(11, r)} = \frac{30}{11} \]

\[ \frac{\frac{10!}{(10-(r+1))!}}{\frac{11!}{(11-r)!}} = \frac{30}{11} \]

\[ \frac{\frac{10!}{(9-r)!}}{\frac{11!}{(11-r)!}} = \frac{30}{11} \]

\[ \frac{10!}{(9-r)!} \times \frac{(11-r)!}{11!} = \frac{30}{11} \]

\[ \frac{10! \times (11-r)!}{11! \times (9-r)!} = \frac{30}{11} \]

\[ \frac{(11-r)(10-r)}{11} = \frac{30}{11} \]

\[ (11-r)(10-r) = 30 \]

\[ 110 - 21r + r^2 = 30 \]

\[ r^2 - 21r + 80 = 0 \]

\[ (r-5)(r-16) = 0 \]

\[ r = 5 \text{ or } r = 16 \]

Since \( P(10, r+1) \) and \( P(11, r) \) must be defined, we need \( r+1 \leq 10 \) and \( r \leq 11 \), so \( r \leq 9 \). Rejecting \( r = 16 \), the answer is \( r = 5 \).
In simple words: Set up the equation from the given condition, expand permutations using factorial notation, cancel common factors, and solve the resulting equation for r.

Exam Tip: Always check that your final answer satisfies the domain constraints - the indices in the permutation notation must be valid. An algebraically correct answer may need to be rejected if it violates these constraints.

Question. Example 6. Prove that
(i) \( P(2n, n) = 2^n [1 \cdot 3 \cdot 5 \cdots (2n-1)] \)
(ii) \( P(n, r) = n \cdot P(n-1, r-1) \)
(iii) \( ^n P_r = ^{n-1} P_r + r \cdot ^{n-1} P_{r-1} \)
Answer:
(i) \( P(2n, n) = \frac{(2n)!}{(2n-n)!} = \frac{(2n)!}{n!} \)

\[ = \frac{1 \cdot 2 \cdot 3 \cdot 4 \cdots (2n-2)(2n-1)(2n)}{1 \cdot 2 \cdot 3 \cdots n} \]

\[ = \frac{[1 \cdot 3 \cdot 5 \cdots (2n-1)] \times [2 \cdot 4 \cdot 6 \cdots (2n)]}{1 \cdot 2 \cdot 3 \cdots n} \]

\[ = \frac{[1 \cdot 3 \cdot 5 \cdots (2n-1)] \times 2^n [1 \cdot 2 \cdot 3 \cdots n]}{1 \cdot 2 \cdot 3 \cdots n} \]

\[ = 2^n [1 \cdot 3 \cdot 5 \cdots (2n-1)] \]

(ii) Right side \( = n \cdot P(n-1, r-1) = n \times \frac{(n-1)!}{(n-1-(r-1))!} = n \times \frac{(n-1)!}{(n-r)!} = \frac{n!}{(n-r)!} = P(n, r) \).

(iii) Right side \( = ^{n-1} P_r + r \cdot ^{n-1} P_{r-1} \)

\[ = \frac{(n-1)!}{(n-1-r)!} + r \times \frac{(n-1)!}{(n-1-(r-1))!} \]

\[ = \frac{(n-1)!}{(n-r-1)!} + r \times \frac{(n-1)!}{(n-r)!} \]

\[ = \frac{(n-1)!}{(n-r-1)!} + \frac{r(n-1)!}{(n-r)(n-r-1)!} \]

\[ = \frac{(n-1)!}{(n-r-1)!} \left( 1 + \frac{r}{n-r} \right) \]

\[ = \frac{(n-1)!}{(n-r-1)!} \times \frac{n-r+r}{n-r} \]

\[ = \frac{(n-1)! \times n}{(n-r)(n-r-1)!} = \frac{n!}{(n-r)!} = ^n P_r \]
In simple words: Use the factorial definition of permutations, separate products into odd and even factors, apply the factorial formula repeatedly, and algebraically simplify to reach the desired result.

Exam Tip: For proofs, show each algebraic step clearly and justify factorizations (e.g., separating (2n)! into odd and even terms). Examiners award marks for clear reasoning, not just the final statement.

Question. Example 7. How many 4 letter words, with or without meaning, can be formed out of the letters of the word WONDER, if repetition of letters is not allowed?
Answer: The word WONDER contains 6 distinct letters. Since repetition is not allowed, any four of these six letters can be arranged in \( ^6 P_4 \) ways.

Hence the required number of 4 letter words \( = ^6 P_4 = 6 \times 5 \times 4 \times 3 = 360 \).
In simple words: We have 6 different letters to choose from, and we pick and arrange 4 of them. For the first letter we have 6 choices, the second has 5 choices, the third has 4, and the fourth has 3 - so the total is 6 × 5 × 4 × 3 = 360.

Exam Tip: Word-formation problems use permutations when repetition is not allowed. The key is to identify how many distinct letters are available and apply the permutation formula \( ^n P_r \).

Question. Example 8. How many 4 letter words, with or without meaning, can be formed out of the letters of the word WONDERFUL, if repetition of letters is not allowed?
Answer: The word WONDERFUL contains 9 distinct letters. Since repetition is not allowed, any four of these nine letters can be arranged in \( ^9 P_4 \) ways.

Hence the required number of 4 letter words \( = 9 \times 8 \times 7 \times 6 = 3024 \).
In simple words: With 9 different letters available and forming 4-letter words, we pick and arrange 4. The first slot has 9 options, second has 8, third has 7, and fourth has 6 - giving 9 × 8 × 7 × 6 = 3024 total words.

Exam Tip: When counting word formations, always verify the number of distinct letters in the source word. Do not count repeated letters multiple times, and always check whether the problem allows repetition.