Basic Concepts of Permutations and Combinations Chapter Notes | Quantitative Aptitude for CA Foundation PDF Download

Chapter Overview

Introduction

• In this chapter, we will explore the challenge of organizing and grouping certain items, focusing on selecting a specific number of items at a time.
• It's important to understand that (a, b) and (b, a) are considered different arrangements, even though they are part of the same group.
• When we talk about arrangements, the order of the items matters.
• For example, a bank manager needs to fill two key roles from a pool of five equally skilled employees.
• Since none of the candidates have prior experience, the manager plans for each one to work in both roles for one month before making a decision.
• This brings up the question: How long can the bank function before these positions are filled permanently?
• The situation described requires careful counting of all possible ways to achieve the desired outcomes.
• While it may be tempting to list all outcomes, doing so can be very time-consuming and prone to mistakes, such as missing or repeating entries.
• To address these challenges, we need to use specific techniques that help in managing such problems.
• The methods of permutation and combination will assist in solving issues like the one presented above.

Fundamental Principles of Counting

(a) Multiplication Rule: If a certain task can be done in 'm' different ways, and after that, a second task can be done in 'n' different ways, then the total number of ways to do both tasks simultaneously is m × n. 

 For example, if there are 5 different buses to go to school and 4 different buses to come back, the total ways to go to and from school is 5 × 4 = 20. 

(b) Addition Rule: If there are two alternative jobs that can be done in 'm' ways and 'n' ways respectively, then either of the two jobs can be done in (m + n) ways. 
For instance, if there are 5 buses and 4 autos to go to school, the total ways to go to school is 5 + 4 = 9. 

Note :- 
(1)

(2) The above fundamental principles may be generalised, wherever necessary.

Formula: The factorial can be expressed as: 

n! = n × (n – 1) × (n – 2) × ... × 3 × 2 × 1

Example 1: Calculate 5!, 4!, and 6!

Solution:
5! = 5 × 4 × 3 × 2 × 1 = 120
4! = 4 × 3 × 2 × 1 = 24
6! = 6 × 5 × 4 × 3 × 2 × 1 = 720

Example 2: Find 9! / 6! and 10! / 7!

Solution:
9! / 6! = 9 × 8 × 7 = 504
10! / 7! = 10 × 9 × 8 = 720

Example 3: Solve for x if 1/9! + 1/10! = x/11!

Solution:
Starting from 1/9! (1 + 1/10) = x/11 × 10 × 9!.
This simplifies to 11/10 = x/11 × 10, leading to x = 121.

Example 4: Find n if

Solution:

Permutations

A group of people wishes to have their photographs taken and approaches a photographer to capture as many different arrangements as possible. The photographer needs to determine how many different films he must use to exhaust all potential arrangements of the individuals standing in various positions. This scenario can be analyzed using permutations, which are used to calculate the arrangements of a set of objects or people, considering the order of arrangement.

To illustrate how permutations work for the photographer, let's consider a specific example with three individuals: Mr. Suresh, Mr. Ramesh, and Mr. Mahesh. The photographer needs to arrange these three individuals in all possible orders. The different arrangements or permutations in this case would be (Suresh, Mahesh, Ramesh), (Suresh, Ramesh, Mahesh), (Ramesh, Suresh, Mahesh), (Ramesh, Mahesh, Suresh), (Mahesh, Ramesh, Suresh), and (Mahesh, Suresh, Ramesh). In total, there are six unique arrangements, indicating that the photographer will need six films to capture all possibilities. Each unique arrangement is referred to as a permutation of three individuals taken at a time.

This may also be exhibited as follows :

with this example as a base, we can introduce a general formula to find the number of permutations.
Number of Permutations when r objects are chosen out of n different objects. (Denoted by ⁿP _r or _nP_r or P_{(n, r)})
Let us consider the problem of finding the number of ways in which the first r rankings are secured by n students in a class. As any one of the n students can secure the first rank, the number of ways in which the first rank is secured is n.
Now consider the second rank. There are (n – 1) students left and the second rank can be secured by any one of them. Thus the different possibilities are (n – 1) ways. Now, applying fundamental principle, we can see that the first two ranks can be secured in n (n – 1) ways by these n students.
After calculating for two ranks, we find that the third rank can be secured by any one of the remaining (n – 2) students. Thus, by applying the generalized fundamental principle, the first three ranks can be secured in n (n – 1) (n – 2) ways .
Continuing in this way we can visualise that the number of ways are reduced by one as the rank is increased by one. Therefore, again, by applying the generalised fundamental principle for r different rankings, we calculate the number of ways in which the first r ranks are secured by n students as

Theorem : The number of permutations of n things when r are chosen at a time

where the product has exactly r factors.

Results

1. Number of permutations of n different things taken all n things at a time is given by

ⁿP_n = n (n – 1) (n – 2) …. (n – n + 1)
=n (n – 1) (n – 2) ….. 2.1 = n!     ...(1)

2. ⁿP _r using factorial notation.

ⁿP_r = n. (n – 1) (n – 2) ….. (n – r + 1)

Thus

3. Justification for 0! = 1. Now applying r = n in the formula for ⁿP_r, we get

ⁿP_n = n!/ (n – n)! = n!/0! 
But from Result 1 we find that ⁿP_n = n!. 
Therefore, by applying this we derive, 0! = n! / n! = 1

Example 1: Evaluate each of ⁵P₃, ¹⁰P₂, ¹¹P₅.

Solution: 

Example 2: How many three letters words can be formed using the letters of the words (a) SQUARE and (b) HEXAGON?
(Any arrangement of letters is called a word even though it may or may not have any meaning or pronunciation).

Solution:
(a) Since the word ‘SQUARE’ consists of 6 different letters, the number of permutations of choosing 3 letters out of six equals
⁶P₃ = 6 × 5 × 4 = 120.
(b) Since the word ‘HEXAGON’ contains 7 different letters, the number of permutations is 
⁷P₃ = 7 × 6 × 5 = 210.

Example 3: In how many different ways can five persons stand in a line for a group photograph?

Solution: Here we know that the order is important. Hence, this is the number of permutations of five things taken all at a time. Therefore, this equals
⁵P₅ = 5! = 5 × 4 × 3 × 2 × 1 = 120 ways.

Example 4: First, second and third prizes are to be awarded at an engineering fair in which 13 exhibits have been entered. In how many different ways can the prizes be awarded?

Solution: Here again, order of selection is important and repetitions are not meaningful as no exhibit can receive more than one prize. Hence , the answer is the number of permutations of 13 things taken three at a time. Therefore, 
we find ¹³P₃ = 13!/10! = 13 × 12 × 11 = 1,716 ways.

Example 5: In how many different ways can 3 students be associated with 4 chartered accountants, assuming that each chartered accountant can take at most one student?

Solution: This equals the number of permutations of choosing 3 persons out of 4. 
Hence , the answer is ⁴P₃ = 4 × 3 × 2 = 24.

Example 6: If six times the number permutations of n things taken 3 at a time is equal to seven times the number of permutations of (n – 1) things taken 3 at a time, find n.

Solution: We are given that 6 × ⁿP₃ = 7 × ^n-1P₃ and we have to solve this equality to find the value of n. Therefore,

or, 6 n (n – 1) (n – 2) = 7 (n – 1) (n – 2) (n – 3) or, 6 n = 7 (n – 3)
or, 6 n = 7n – 21
or, n = 21
Therefore, the value of n equals 21.

Example 7: Compute the sum of 4 digit numbers which can be formed with the four digits 1, 3, 5, 7, if each digit is used only once in each arrangement.

Solution: The number of arrangements of 4 different digits taken 4 at a time is given by ⁴P₄ = 4! = 24. All the four digits will occur equal number of times at each of the positions, namely ones, tens, hundreds, thousands.

Thus, each digit will occur 24 / 4 = 6 times in each of the positions. The sum of digits in one’s position will be 6 × (1 + 3 + 5 + 7) = 96. Similar is the case in ten’s, hundred’s and thousand’s places. Therefore, the sum will be 96 + 96 × 10 + 96 × 100 + 96 × 1000 = 1,06,656.

Example 8: Find n if ⁿP₃ = 60.

Solution:

i.e., n (n–1) (n–2) = 60 = 5 × 4 × 3 
Therefore, n = 5.

Example 9: If ⁵⁶P_r+6 : ⁵⁴P _r+3 = 30,800 : 1, find r.

Solution: We know

But we are given the ratio as 30800 : 1 ; therefore

Example 10: Prove the following 

Solution: By applying the simple properties of factorial, we have (n +1)! – n! = (n+1) n! – n! = n!. (n+1–1) = n. n!

Example 11: In how many different ways can a club with 10 members select a President, Secretary and Treasurer, if no member can hold two offices and each member is eligible for any office?

Solution: The answer is the number of permutations of 10 persons chosen three at a time. Therefore, it is ¹⁰p₃ = 10 × 9 × 8 = 720.

Example 12: When Jhon arrives in New York, he has eight shops to see, but he has time only to visit six of them. In how many different ways can he arrange his schedule in New York?

Solution: He can arrange his schedule in ⁸P₆ = 8 × 7 × 6 × 5 × 4 × 3 = 20,160 ways.

Example 13: When Dr. Ram arrives in his dispensary, he finds 12 patients waiting to see him. If he can see only one patient at a time, find the number of ways, he can schedule his patients (a) if they all want their turn, and (b) if 3 leave in disgust before Dr. Ram gets around to seeing them.

Solution: (a) There are 12 patients and all 12 wait to see the doctor. Therefore the number of ways = ¹²P₁₂ = 12! = 479,001,600
(b) There are 12 – 3 = 9 patients. They can be seen ¹²P₉ = 79,833,600 ways.

Circular Permutations

The number of circular permutations of n different things chosen at a time is (n–1)!.

Proof : Let any one of the permutations of n different things taken. Then consider the rearrangement of this permutation by putting the last thing as the first thing. Even though this is a different permutation in the ordinary sense, it will not be different in all n things are arranged in a circle. Similarly, we can consider shifting the last two things to the front and so on. Specially, it can be better understood, if we consider a,b,c,d. If we place a,b,c,d in order, then we also get abcd, dabc, cdab, bcda as four ordinary permutations. These four words in circular case are one and same thing. See below circles.

Thus we find in above illustration that four ordinary permutations equals one in circular.
Therefore, n ordinary permutations equal one circular permutation.
Hence there are ⁿP_n/ n ways in which all the n things can be arranged in a circle. This equals (n– 1)!.

Example 1: In how many ways can 4 persons sit at a round table for a group discussions?

Solution: The answer can be get from the formula for circular permutations. The answer is (4–1)! = 3! = 6 ways.

NOTE : These arrangements are such that every person has got the same two neighbours. The only change is that right side neighbour and vice-versa.

Thus the number of ways of arranging n persons along a round table so that no person has the same two neighbours is
Similarly, in forming a necklace or a garland there is no distinction between a clockwise and anti clockwise direction because we can simply turn it over so that clockwise becomes anti clockwise and vice versa. Hence, the number of necklaces formed with n beads of different colours

Permutations with Restrictions

In many arrangements there may be number of restrictions. in such cases, we are to arrange or select the objects or persons as per the restrictions imposed. The total number of arrangements in all cases, can be found out by the application of fundamental principle.

Theorem 1. Number of permutations of n distinct objects taken r at a time when a particular object is not taken in any arrangement is ^n–1p_r.

Proof : Since a particular object is always to be excluded, we have to place n – 1 objects at r places. Clearly this can be done in ^n–1p_r ways.

Theorem 2. Number of permutations of r objects out of n distinct objects when a particular object is always included in any arrangement is r. ^n-1p_r-1

Proof : If the particular object is placed at first place, remaining r – 1 places can be filled from n – 1 distinct objects in ^n–1p_r–1 ways. Similarly, by placing the particular object in 2nd, 3rd, ….., rth place, we find that in each case the number of permutations is ^n–1p_r–1.This the total number of arrangements in which a particular object always occurs is r. ^n–1p_r–1

The following examples will enlighten further:

Example 1: How many arrangements can be made out of the letters of the word `DRAUGHT’, the vowels never beings separated?

Solution: The word `DRAUGHT’ consists of 7 letters of which 5 are consonants and two are vowels. In the arrangement we are to take all the 7 letters but the restriction is that the two vowels should not be separated.
We can view the two vowels as one letter. The two vowels A and U in this one letter can be arranged in 2! = 2 ways. (i) AU or (ii) UA. Further, we can arrange the six letters : 5 consonants and one letter (compound letter consisting of two vowels). The total number of ways of arranging them is ⁶P₆ = 6! = 720 ways.
Hence, by the fundamental principle, the total number of arrangements of the letters of the word DRAUGHT, the vowels never being separated = 2 × 720 = 1440 ways.

Example 2: Show that the number of ways in which n books can be arranged on a shelf so that two particular books are not together.The number is (n–2).(n–1)!

Solution: We first find the total number of arrangements in which all n books can be arranged on the shelf without any restriction. The number is,ⁿP_n = n! ….. (1)
Then we find the total number of arrangements in which the two particular books are together.
The books can be together in ²P₂ = 2! = 2 ways. Now we consider those two books which are kept together as one composite book and with the rest of the (n–2) books from (n–1) books which are to be arranged on the shelf; the number of arrangements = ^n–1P_n–1 = (n–1) !. Hence by the Fundamental Principle, the total number of arrangements on which the two particular books are together equals = 2 × (n–1)! …….(2) the required number of arrangements of n books on a shelf so that two particular books are not together
= (1) – (2)
= n! – 2 x (n–1)!
= n.(n – 1)! – 2 . (n–1)!
= (n–1)! . (n–2)

Example 3: There are 6 books on Economics, 3 on Mathematics and 2 on Accountancy. In how many ways can these be placed on a shelf if the books on the same subject are to be together?

Solution: Consider one such arrangement. 6 Economics books can be arranged among themselves in 6! Ways, 3 Mathematics books can be arranged in 3! Ways and the 2 books on Accountancy can be arranged in 2! ways. Consider the books on each subject as one unit. Now there are three units. These 3 units can be arranged in 3! Ways.
Total number of arrangements
= 3! × 6! × 3! × 2!
= 51,840.

Example 4: How many different numbers can be formed by using any three out of five digits 1, 2, 3, 4, 5, no digit being repeated in any number?
How many of these will (i) begin with a specified digit? (ii) begin with a specified digit and end with another specified digit?

Solution: Here we have 5 different digits and we have to find out the number of permutations of them 3 at a time. Required number is ⁵P₃ = 5.4.3 = 60.

(i) If the numbers begin with a specified digit, then we have to find the number of Permutations of the remaining 4 digits taken 2 at a time. Thus, desired number is ⁴P₂ = 4.3 = 12.

(ii) Here two digits are fixed; first and last; hence, we are left with the choice of finding the number of permutations of 3 things taken one at a time i.e., ³P₁ =3.

Example 5: How many four digit numbers can be formed out of the digits 1,2,3,5,7,8,9, if no digit is repeated in any number? How many of these will be greater than 3000?

Solution: We are given 7 different digits and a four-digit number is to be formed using any 4 of these digits. This is same as the permutations of 7 different things taken 4 at a time.
Hence, the number of four-digit numbers that can be formed = ⁷P₄ = 7 × 6 × 5 × 4 × = 840 ways.
Next, there is the restriction that the four-digit numbers so formed must be greater than 3,000. Thus, it will be so if the first digit-that in the thousand’s position, is one of the five digits 3, 5, 7, 8, 9. Hence, the first digit can be chosen in 5 different ways; when this is done, the rest of the 3 digits are to be chosen from the rest of the 6 digits without any restriction and this can be done in ⁶P₃ ways.
Hence, by the Fundamental principle, we have the number of four-digit numbers greater than 3,000 that can be formed by taking 4 digits from the given 7 digits = 5 × ⁶P₃ = 5 × 6 × 5 × 4 = 5 × 120 = 600.

Example 6: Find the total number of numbers greater than 2000 that can be formed with the digits 1, 2, 3, 4, 5 no digit being repeated in any number.

Solution: All the 5 digit numbers that can be formed with the given 5 digits are greater than 2000. This can be done in

⁵P₅ = 5! = 120 ways …...................................(1)
The four digited numbers that can be formed with any four of the given 5 digits are greater than 2000 if the first digit, i.e.,the digit in the thousand’s position is one of the four digits 2, 3, 4, 5. this can be done in ⁴P₁ = 4 ways. When this is done, the rest of the 3 digits are to be chosen from the rest of 5–1 = 4 digits. This can be done in ⁴P₃ = 4 × 3 × 2 = 24 ways.
Therefore, by the Fundamental principle, the number of four-digit numbers greater than 2000 = 4 × 24 = 96 …. (2)
Adding (1) and (2), we find the total number greater than 2000 to be 120 + 96 = 216.

Example 7: There are 6 students of whom 2 are Indians, 2 Americans, and the remaining 2 are Russians. They have to stand in a row for a photograph so that the two Indians are together, the two Americans are together and so also the two Russians. Find the number of ways in which they can do so.

Solution: The two Indians can stand together in ²P₂ = 2! = 2 ways. So is the case with the two Americans and the two Russians.
Now these 3 groups of 2 each can stand in a row in ³P₃ = 3 x 2 = 6 ways. Hence by the generalized fundamental principle, the total number of ways in which they can stand for a photograph under given conditions is  6 × 2 × 2 × 2 = 48

Example 8: A family of 4 brothers and three sisters is to be arranged for a photograph in one row. In how many ways can they be seated if (i) all the sisters sit together, (ii) no two sisters sit together?

Solution: (i) Consider the sisters as one unit and each brother as one unit. 4 brothers and 3 sisters make 5 units which can be arranged in 5! ways. Again 3 sisters may be arranged amongst themselves in 3! Ways
Therefore, total number of ways in which all the sisters sit together = 5!×3! = 720 ways.
(ii) In this case, each sister must sit on each side of the brothers. There are 5 such positions as indicated below by upward arrows :

4 brothers may be arranged among themselves in 4! ways. For each of these arrangements 3 sisters can sit in the 5 places in 5P3 ways.
Thus the total number of ways = ⁵P₃ × 4! = 60 × 24 = 1,440

Example 9: In how many ways can 8 persons be seated at a round table? In how many cases will 2 particular persons sit together?

Solution: This is in form of circular permutation. Hence the number of ways in which eight persons can be seated at a round table is ( n – 1 )! = ( 8 – 1 )! = 7! = 5040 ways.
Consider the two particular persons as one person. Then the group of 8 persons becomes a group of 7 (with the restriction that the two particular persons be together) and seven persons can be arranged in a circular in 6! Ways.

Hence, by the fundamental principle, we have, the total number of cases in which 2 particular persons sit together in a circular arrangement of 8 persons = 2! × 6! = 2 × 6 × 5 × 4 × 3 × 2 × 1 = 1,440.

Example 10: Six boys and five girls are to be seated for a photograph in a row such that no two girls sit together and no two boys sit together. Find the number of ways in which this can be done.

Solution: Suppose that we have 11 chairs in a row and we want the 6 boys and 5 girls to be seated such that no two girls and no two boys are together. If we number the chairs from left to right, the arrangement will be possible if and only if boys occupy the odd places and girls occupy the even places in the row. The six odd places from 1 to 11 may filled in by 6 boys in 6P6 ways. Similarly, the five even places from 2 to 10 may be filled in by 5 girls in ⁵P₅ ways.

Hence, by the fundamental principle, the total number of required arrangements = ⁶P₆ × ⁵P₅ = 6! × 5! = 720 × 120 = 86,400.

Combinations

A combination is defined as the number of ways to arrange or select a subset of items from a larger set, where the order of selection or arrangement is not important. A practical example of this is the selection of a poker hand, which consists of five cards chosen from a total of 52 cards. This illustrates the idea of combinations: choosing 5 items from a set of 52.

The number of combinations of n different items taken r at a time is denoted as nC_r (or C(n, r) or C_rⁿ). In this notation, nC_r represents the required number of combinations. Each combination consists of r items, and since the order does not matter, we are not considering permutations. If we were to consider order, we would be dealing with permutations of r items from n. Thus, for every combination of r items, there are rP_r permutations possible. Therefore, the relationship can be summarized as nC_r × rP_r = nP_r, indicating that the total combinations multiplied by their permutations equals the total permutations of r items chosen from n.
The combination formula, denoted as nC_r, is derived from the permutation formula and is expressed as:
nC_r = nP_r / rP_r
This can be further simplified using the definitions of permutations:
nC_r = nP_r / rP_r = n! / (n - r)! ÷ r! / (r - r)! = n! / (n - r)! × 0! / r!
Consequently, we arrive at the formula:
nC_r = n! / (r! (n - r)!)
From this formula, we can derive some important results:
(i) For nC₀, substituting r = 0 yields:
nC₀ = n! / (0! (n - 0)!) = n! / n! = 1 (since 0! = 1)
(ii) For nC_n, substituting r = n gives:
nC_n = n! / (n! (n - n)!) = n! / (n! 0!) = 1
This demonstrates that there is exactly one way to choose all or none from a set.

Example 1: Find the number of different poker hands in a pack of 52 playing cards.

Solution: This is the number of combinations of 52 cards taken five at a time. Now applying the formula,

Example 2: Let S be the collection of eight points in the plane with no three points on the straight line. Find the number of triangles that have points of S as vertices.

Solution: Every choice of three points out of S determines a unique triangle. The order of the points selected is unimportant as whatever be the order, we will get the same triangle. Hence, the desired number is the number of combinations of eight things taken three at a time. Therefore, we get

⁸C₃ = 8!/3!5! = 8 × 7 × 6/3 × 2 × 1 = 56 choices.

Example 3: A committee is to be formed of 3 persons out of 12. Find the number of ways of forming such a committee.

Solution: We want to find out the number of combinations of 12 things taken 3 at a time and this is given by

¹²C₃ = 12!/3!(12 – 3)! [ by the definition of ⁿC_r] 
= 12!/3!9! = 12 × 11 × 10 × 9!/3!9! = 12 × 11 × 10/3 × 2 = 220

Example 4: A committee of 7 members is to be chosen from 6 Chartered Accountants, 4 Economists and 5 Cost Accountants. In how many ways can this be done if in the committee, there must be at least one member from each group and at least 3 Chartered Accountants?

Solution: The various methods of selecting the persons from the various groups are shown below:

Number of ways of choosing the committee members by

Therefore, total number of ways = 1,200 + 450 + 600 + 120 + 400 + 800 = 3,570

Example 5: A person has 12 friends of whom 8 are relatives. In how many ways can he invite 7 guests such that 5 of them are relatives?

Solution: Of the 12 friends, 8 are relatives and the remaining 4 are not relatives. He has to invite 5 relatives and 2 friends as his guests. 5 relatives can be chosen out of 8 in ⁸C₅ ways; 2 friends can be chosen out of 4 in ⁴C₂ ways.
Hence, by the fundamental principle, the number of ways in which he can invite 7 guests such that 5 of them are relatives and 2 are friends.

= ⁸C₅ × ⁴C₂

= 336.

Example 6: A Company wishes to simultaneously promote two of its 6 department heads to assistant managers. In how many ways these promotions can take place?

Solution: This is a problem of combination. Hence, the promotions can be done in

⁶C₂ = 6×5 / 2 = 15 ways

Example 7: A building contractor needs three helpers and ten men apply. In how many ways can these selections take place?

Solution: There is no regard for order in this problem. Hence, the contractor can select in any of ¹⁰C₃ ways i.e.,

(10 × 9 × 8) / (3 × 2 × 1) = 120 ways.

Example 8: In each case, find n:

Solution: (a) 4. ⁿC₂ = ⁿ⁺²C₃ (b) ⁿ⁺²C_n = 45.

(a) We are given that 4. ⁿC₂ = ⁿ⁺²C₃. Now applying the formula,

(b) We are given that ⁿ⁺²C_n = 45. Applying the formula,

Thus, n equals either – 11 or 8. But negative value is not possible. Therefore we conclude that n=8.

Example 9: A box contains 7 red, 6 white and 4 blue balls. How many selections of three balls can be made so that (a) all three are red (b) none is red (c) one is of each colour?

Solution: (a) All three balls will be of red colour if they are taken out of 7 red balls and this can be done in

⁷C₃ = 7! / 3!(7–3)!

= 7! / 3!4! = 7 × 6 × 5 × 4! / (3 × 2 × 4!) 
= 7 × 6 × 5 / (3 × 2) = 35 ways

Hence, 35 selections (groups) will be there such that all three balls are red.

(b) None of the three will be red if these are chosen from (6 white and 4 blue balls) 10 balls and this can be done in

¹⁰C₃ = 10!/{3!(10–3)!} = 10! / 3!7! 
= 10 × 9 × 8×7! / (3 × 2 × 1 × 7!) = 10 × 9 × 8 /(3 × 2) = 120 ways.

Hence, the selections (or groups) of three such that none is a red ball are 120 in number.

One red ball can be chosen from 7 balls in ⁷C₁ = 7 ways. One white ball can be chosen from 6 white balls in ⁶C₁ ways. One blue ball can be chosen from 4 blue balls in ⁴C₁ = 4 ways. Hence, by generalized fundamental principle, the number of groups of three balls such that one is of each colour = 7 × 6 × 4 = 168 ways.

Example 10: If ¹⁰P_r = 6,04,800 and ¹⁰C _r = 120; find the value of r,

Solution: We know that nC r. rP r =nP r. We will us this equality to find r.

¹⁰P_r = ¹⁰C_r .r!

or, 6,04,800 =120 ×r!
or, r! = 6,04,800 ÷120 = 5,040
But r! = 5040 = 7×6×4×3×2×1 = 7!
Therefore, r=7.

Properties of nC r :

Example 11: Find r if ¹⁸C_r = ¹⁸C _r+2

Solution: We know that ⁿC_r is equal to ⁿC _(n−r). Therefore, we can say: 

¹⁸C _r = ¹⁸C _(18−r)

 It is given that ¹⁸C _r is equal to ¹⁸C _r+2. So we have: 

¹⁸C _(18−r) = ¹⁸C _r+2

 This leads us to the equation: 

18 − r = r + 2

 By rearranging the equation, we can solve for r: 

2r = 18 − 2 = 16

 Thus, we find: 

r = 8

Example 12: Prove that 

Solution: R.H.S 

Hence, the result

Example 13: If ²⁸C_2r : ²⁴C_2r–4 = 225 : 11, find r.

Solution: We have ⁿC_r = n! / {r!(n–r)!} Now, substituting for n and r, we get

= 11 × 28 × 3 × 26

= 11 × 7 × 4 × 3 × 13 × 2

= 11 × 12 × 13 × 14

= 14 × 13 ×12 x 11

∴2r= 14 i.e., r = 7

Example 14: Find x if

Solution: 

L.H.S 

Also ⁿC _r =ⁿC_n–r. Therefore ¹⁴C ₅ = ¹⁴C_14–5 = ¹⁴C₉

Hence,  by the given equality
This implies, either x = 5 or x = 9.

Standard Results

 Now, let's explore some standard results in permutations and combinations that have special applications. 

I. Permutations with Identical Items

•  When arranging n things where n1 are identical of one kind, n2 are identical of another kind, n3 are identical of a third kind, and the rest are different, the number of ways (p) to arrange them is given by: 
• p = n! / (n1! n2! n3!)
•  This formula can be extended to cases with different groups of identical items. 

II. Permutations with Repetition

•  The number of permutations of n things taken r at a time, where each thing can be repeated r times, is given by: nr
•  This is because for each position, we have n choices, and this is true for all r positions. 

III. Combinations of Different Things

•  The total number of ways to form groups by taking some or all of n different things at a time is: 2ⁿ – 1
•  This is because each of the n different things can be either included or excluded, resulting in 2n combinations. 

IV. Combinations with Identical Items

•  The total number of ways to form groups by taking some or all out of n things, where n1 are identical of one kind, n2 are identical of another kind, and so on, is given by: (n₁ + 1)(n₂ + 1)(n₃ + 1) – 1
•  This formula accounts for the different ways to include the identical items and excludes the case where none are taken. 

V. Independence in Combinations

•  Combinations are considered independent when selecting subsets from different sets does not affect each other. 
•  The total number of combinations can be calculated by multiplying the combinations from each set:

Note : This result can be extended to more than two sets of objects by a similar reasoning.

Example 1: How many different permutations are possible from the letters of the word 'CALCULUS’?

Solution: The word `CALCULUS’ consists of 8 letters of which 2 are C and 2 are L, 2 are U and the rest are A and S. Hence , by result (I), the number of different permutations from the letters of the word `CALCULUS’ taken all at a time =

Example 2: In how many ways can 17 billiard balls be arranged , if 7 of them are black, 6 red and 4 white?

Solution: We have, the required number of different arrangements:

Example 3: An examination paper with 10 questions consists of 6 questions in Algebra and 4 questions in Geometry. At least one question from each section is to be attempted. In how many ways can this be done?

Solution: A student must answer atleast one question from each section and he may answer all questions from each section.

Consider Section I : Algebra. There are 6 questions and he may answer a question or may not answer it. These are the two alternatives associated with each of the six questions. Hence, by the generalised fundamental principle, he can deal with two questions in 2 × 2 ….6 factors = 2⁶ number of ways. But this includes the possibility of none of the question from Algebra being attempted. This cannot be so, as he must attempt at least one question from this section. Hence, excluding this case, the number of ways in which Section I can be dealt with is (2⁶ –1).

Similarly, the number of ways in which Section II can be dealt with is (2⁴ –1).

Hence, by the Fundamental Principle, the examination paper can be attempted in (2⁶ –1) (2⁴ –1) number of ways.

Example 4: A man has 5 friends. In how many ways can he invite one or more of his friends to dinner?

Solution: By result, (III) of this section, as he has to select one or more of his 5 friends, he can do so in 2⁵ – 1 = 31 ways.

Note : This can also be done in the way, outlines below. He can invite his friends one by one, in twos, in threes, etc. and hence the number of ways.

Example 5: There are 7 men and 3 ladies. Find the number of ways in which a committee of 6 can be formed of them if the committee is to include atleast two ladies?

Solution: The committee of six must include at least 2 ladies, i.e., two or more ladies. As there are only 3 ladies, the following possibilities arise:

The committee of 6 consists of (i) 4 men and 2 ladies (ii) 3 men and 3 ladies.

The number of ways for (i) = ⁷C₄ × ³C₂ = 35 × 3 = 105;

The number of ways for (ii) = ⁷C₃ × ³C₃ = 35 × 1 = 35.

Hence the total number of ways of forming a committee so as to include at least two ladies = 105 +35 = 140.

Example 6: Find the number of ways of selecting 4 letters from the word `EXAMINATION’.

Solution: There are 11 letters in the word of which A, I, N are repeated twice.
Thus we have 11 letters of 8 different kinds (A, A), (I, I), (N, N), E, X, M, T, O.
The group of four selected letters may take any of the following forms:
(i) Two alike and other two alike
(ii) Two alike and other two different
(iii) All four different
In case (i), the number of ways = ³C₂ = 3.
In case (ii), the number of ways = ³C₁ × ⁷C₂ = 3 × 21 = 63.
In case (iii), the number of ways =

Hence , the required number of ways = 3 + 63 + 70 = 136 ways