Permutations and Combinations Formulas: Derivation, Example

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Permutations and Combinations are methods used to express a group of elements by arranging them in a set and forming subsets. Permutations and Combinations Formulas are used to arrange certain groups of data in a specific order.

Permutation is a method of arranging all the elements of a set into a sequence or order.
Combination is a method of selecting elements from a collection, such that the order of selection does not matter.
Permutations can be considered as ‘Arrangements’ and combinations can be referred to as ‘Selections’.

Permutation Formula is used to arrange r things from a set of n things without replacement. Permutation Formula is given as:

\({^nP_r} = \frac{n!}{(n-r)!} \)

Combination Formula is used to choose r things from a set of n things without replacement. Combination Formula is given as:

\({^nC_r} = \frac{n!}{r!(n-r)!} \)

Table of Content

What is Permutation?
What is Combination?
Permutation and Combination Formulas
Derivation of Permutation and Combination Formulas
Things to Remember
Previous Years’ Questions
Sample Questions

Key Terms: Permutations and Combinations Formula, Permutations, Combinations, Permutation Formula, Combination Formula

What is Permutation?

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Permutation is defined as the way of arranging all the elements of a set into a finite order or sequence.

In simpler words, Permutation is the sorting of elements of a set in ascending order or descending order.
If the set of elements is already in sorted order, then rearranging elements of the set is known as Permuting.
It is used to determine all possible arrangement sequences of a single data point.
The sequence or order of arrangement matters in Permutation.
Permutations are used in almost every field and sector of Mathematics.

Permutations

Permutations

Permutation Examples

Some examples of Permutation are as follows:

Arranging people, digits, numbers, alphabets, letters, etc.
Picking a team captain, pitcher, and goalkeeper from a group.
Picking first, second, and third place winners.

Permutations and Combinations Detailed Video Explanation

What is Combination?

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Combination is defined as a way of selecting elements from a set, in which the order of selection doesn’t matter.

It is the combination of n things taken k at a time without any repetition.
The order or sequence of arrangement doesn’t matter in Combinations.
In smaller cases, it is possible to count the number of combinations manually, however, it needs a formula for tedious calculations.
Combination is also used in every field of mathematics just like Permutation.

Combinations

Combinations

Combination Examples

Some examples of Combination are as follows:

Picking up team members from a group
Picking three winners.
Picking three colors from a color brochure.

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Relevant Concepts
Combinatorics	Sets Formula	Difference between Permutation and Combination
Set Theory in Maths	Factorial Formula	Relations and Functions

Permutation and Combination Formulas

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Permutation and Combination Formulas are the two key formulas used in basic methods of counting in Mathematics. They are used to find out the permutations and combinations of r items taken from n items.

Permutation is a method used to find the different arrangements of elements of a set.
Combination is used to find the different groups from a larger set.
Permutation is always greater than the combination for any given values of n and r.

Permutation Formula

Permutation Formula is used is used to arrange elements of a set in sequence or order. When repetition is not allowed, Permutation Formula is given as follows:

\({^nP_r} = \frac{n!}{(n-r)!} \)

Where

n refers to the Total number of objects in a given set.
r refers to the Number of objects to be chosen from a given set.

Permutation with repetition of items can also be easily calculated using the exponent form. Thus, to find the number of permutations of n different things, taken r at a time, where repetition is allowed, the Permutation Formula is

P (n, r) = n^r

Thus, Permutation Formula is used to arrange elements of a set with or without repetition.

Solved Example

Example: What will be the number of 3-letter words that can be formed from the letters a, b, c, d, and e in which the letters are not allowed to be repeated?

Solution: It is given that,

Number of Letters Given, n = 5
Number of Letters in Each Word, r = 3

Using the Permutation Formula,

\({^nP_r} = \frac{n!}{(n-r)!} \)

Substituting the values, we get

⁵P₃ = (5!) / (5 - 3)! = 5! / 2! = 120/2 = 60

So, there are 60 different permutations to form a 3-letter word from the letters a, b, c, d, and e in which the letters are not allowed to be repeated.

Combination Formula

Combination is a method of choosing a subset from a larger set of things where the order or sequence doesn't matter. Combination Formula is given as follows:

\({^nC_r} = \frac{{^nP_r}}{r!} = \frac{n!}{r!(n-r)!} \)

Where

n refers to the Total number of objects in a given set.
r refers to the Number of objects to be chosen from the given set.

Solved Example

Example: Find the number of ways can a committee be formed of 5 men and 3 women picked from a group of 9 men and 12 women.

Solution: According to the question, the committee is to be made of a combination of 5 men and 3 women.

Using the Combination Formula,

\({^nC_r} = \frac{n!}{r!(n-r)!} \)

Here, assume n being the total number of men to choose from and r be the number of men to be chosen.

⁹C₅ = \( = \frac{9!}{5!(4)!} = 126\)

Thus, 5 men can be chosen out of 9 men in 126 ways.

Similarly, in ⁿC_r, with n being the total number of women and p being the number of women to be chosen.

¹²C₃ = \( = \frac{12!}{3!(9)!} = 220\)

Thus, 3 women can be chosen out of 12 women in 220 ways.

Hence, the committee can be chosen in 126 × 220 = 27720 ways.

Derivation of Permutation and Combination Formulas

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Permutation and Combination Formulas are used to express a group of objects by arranging them in a certain order or forming their subsets.

Permutation Formula is given as \({^nP_r} = \frac{n!}{(n-r)!} \).
Combination Formula is given as \({^nC_r} = \frac{n!}{r!(n-r)!} \).

Given below is the detailed derivation of Permutation and Combination Formulas.

Derivation of Permutation Formula

Consider that there are r cartons and each of them can hold only one thing. Thus, there will be as many permutations as there are ways of filling in r vacant cartons by n objects.

Number of Ways First Box can be Filled: n
Number of Ways Second Box can be Filled: (n – 1)
Number of Ways Third Box can be Filled: (n – 2)
Number of Ways Fourth Box can be Filled: (n – 3)
Number of Ways r^th Box can be Filled: (n – (r – 1))

Thus, the number of ways of filling in r boxes in succession is given by:

n (n – 1) (n – 2) (n-3) . . . (n – (r – 1))

It can also be written as:

n (n – 1) (n – 2) … (n – r + 1)

The number of permutations of n different objects taken r at a time, where the objects do not repeat is:

n (n – 1) (n – 2) (n – 3) . . . (n – r + 1)

ⁿP_r = n ( n – 1) ( n – 2)( n – 3). . .( n – r + 1) …….. (1)

On multiplying and dividing (1) by (n-r) (n-r-1) (n-r-2)........... 3.2.1, we get

\(P (n, r) =\frac{n.(n-1).(n-2).…. (n-r+1)[(n-r) (n-r-1) (n-r-2)... 3. 2. 1]}{[(n-r) (n-r-1) (n-r-2)....3. 2. 1]}\)

\(p (n, r)= \frac{n!}{(n-r)!}\)

Hence Derived.

Derivation of Combination Formula

Combination is defined as the number of combinations of n distinct things taking r at a time where the order doesn't matter. Now,

Number of Ways to Select First Object from n Distinct Objects: n Ways
Number of Ways to Select Second Object from (n-1) Distinct Objects: (n-1) Ways
Number of Ways to Select Third Object from (n-2) Distinct Objects:(n-2) Ways
Number of Ways to Select Fourth Object from (n-3) Distinct Objects: (n-3) Ways
Number of Ways to Select r^th Object from (n-(r-1)) Distinct Objects: (n-(r-1)) Ways

An ordered subset of r elements is made by completing the selection of r things of the original set of n things. Thus, the number of ways to make a selection of r elements of the original set of n elements will be

n (n – 1) (n – 2) (n-3) . . . (n – (r – 1)) or n (n – 1) (n – 2) … (n – r + 1)

Now, consider the ordered subset of r elements and all its permutations. The total number of all permutations of the given subset will be equal to r! as r objects in every combination can be rearranged in r! ways. Thus,

C(n,r) = P (n, r)/ r!

\(C(n,r) =\dfrac{\frac{n!}{(n-r)!}}{r!}\)

Therefore, the Combination Formula is derived as:

\(C(n, r) = \dfrac{n!}{r!.(n - r)!}\)

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Related Topics
Calculus	Union of Sets	Probability and Statistics
Types of Sets	Fundamental Theorem of Arithmetic	Sequence and Series

Things to Remember

Permutations and Combinations are the counting methods used to determine the number of possible outcomes.
Permutation is a way of arranging all the elements of a set into a finite sequence or order.
Permutation Formula is expressed as \({^nP_r} = \frac{n!}{(n-r)!} \).
Combination is a way of forming subsets by selecting data from a larger set.
Combination Formula is expressed as \({^nC_r} = \frac{n!}{r!(n-r)!} \).
The order or sequence of elements doesn’t matter in Combination, unlike Permutation.

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Previous Years’ Questions

Sample Questions

Ques. (a) Find the number of words that can be formed using the letters of the word “TRIANGLES”.
(b) Determine how many of these words start with “T” and end with “S”. (3 Marks)

Ans. (a) There are 9 letters in the given word which are not repeated in the whole string. Thus, the number of permutations of the letters of the word is ⁹P₉ is,

Using the Permutations Formula,

\({^nP_r} = \frac{n!}{(n-r)!} \)

⁹P₉ = (9!) / (9 - 9)! = (9!) / (0!) { 0! factorial is 1}

⁹P₉ = 9!

So, there are 9! ways in which letters of the word “TRIANGLES” can be formed.

(b) If we put “T” at the start and “S” at the end of the world, then we have to put 7 different letters in 7 places. This can be done in ⁷P₇.

Using the Permutations Formula,

\({^nP_r} = \frac{n!}{(n-r)!} \)

⁷P₇= (7!) / (7 - 7)! = (7!) / (0!) {0! factorial is 1}

⁷P₇= 7!

Thus, there are 7! words that start with “T” and end with “S.

Ques. A permutation lock will open if the correct choice of 3 numbers from 1 to 50 is selected. Determine how many lock permutations can be made assuming no number is repeated. (3 Marks)

Ans. There are a total of 50 digits out of which 3 digits need to be arranged to get the right lock. Thus, there is a possibility of ⁵⁰P₃ ways.

Using the Permutations Formula,

\({^nP_r} = \frac{n!}{(n-r)!} \)

⁵⁰P₃ = (50!) / (50 - 3)! = (50!) / (47!)

If we open the factorial in the numerator till 47! then 47! in the numerator gets canceled with 47! in the denominator.

= (50 x 49 x 48 x 47!) / (47!)

= 50 x 49 x 48 = 117, 600 Ways

Thus, there are 117, 600 possible ways to open the permutation lock.

Ques. There are six friends that want to play enough games of chess to be sure everyone plays everyone else. How many games will they have to play? (3 Marks)

Ans. There are six players in total and two players are to be taken at a time for the game.

Using the Combination Formula,

\({^nC_r} = \frac{{^nP_r}}{r!} = \frac{n!}{r!(n-r)!} \)

C (6, 2) = ⁶C₂ = 6! / 4! x 2! = 6 x 5 x 4! / 4! x 2!

4! in the numerator and 4! in the denominator will be canceled out.

= 30 / 2x1 = 30 / 2 = 15

Thus, the players need to play 15 games in total.

Ques. In a lottery, each ticket has 5 one-digit numbers 0 – 9 written on it.
(a) One win if the ticket has the digits in any order. What are the chances of winning the game?
(b) One would win only if the ticket has the digits in the required order. What are the chances of winning? (5 Marks)

Ans. There are 10 digits in total to be taken 5 at a time.

(a) Using the Combination Formula,

\({^nC_r} = \frac{{^nP_r}}{r!} = \frac{n!}{r!(n-r)!} \)

C (10, 5) = ¹⁰C₅ = 10! / 5! * 5!

= 10 x 9 x 8 x 7 x 6 x 5! / 5! x 5!

5! in the numerator and 5! in the denominator will be canceled out.

= 10 x 9 x 8 x 7 x 6 / 5 x 4 x 3 x 2 x 1

= 252.

Thus, the chances of winning are 1 out of 252.

(b) In this case order of elements matters. So, we will be going to use permutations in this case.

Using the Permutations Formula,

\({^nP_r} = \frac{n!}{(n-r)!} \)

P (10, 5) = ¹⁰P₅ = (10!) / (5!) = 10 * 9 * 8 * 7 * 6 * 5! / 5!

5! in the numerator and 5! in the denominator will be canceled out.

= 10 x 9 x 8 x 7 x 6 = 30240 Ways

So, the chance of winning, in this case, is 30240.

Ques. If the letters of the word RACHIT are arranged in all possible ways as listed in a dictionary, then, determine the rank of the word RACHIT. (3 Marks)

Ans. The alphabetical order of the letters of the word RACHIT is A, C, H, I, R, T.

Number of Words beginning with A = 5!
Number of Words beginning with C = 5!
Number of Words beginning with H = 5!
Number of Words beginning with I = 5!

Clearly, the first word beginning with R is RACHIT.

Thus, the Rank of the word RACHIT in the dictionary = 4×5! + 1= 4 x 120+1= 481

Ques. A candidate needs to answer 7 questions out of 12 questions, which are divided into two groups, each containing 6 questions. The candidate is not permitted to attempt more than 5 questions from either group. What will be the number of different ways of attempting the questions? (3 Marks)

Ans. The candidate cannot attempt more than 5 questions from either group, thus, he is able to attempt a minimum of two questions from either group.

The possible number of questions attempted by each group will be as given in the following table:

Group I	5	4	3	2
Group II	2	3	4	5

Therefore, the total number of possible ways = 2 [⁶C₅ X ⁶C₂ + ⁶C₄ X ⁶C₃]

= 2 [6 x 15 + 15 x 20]

= 2 x 390

= 780

Ques. A company wishes to select 6 persons from 8 but, if person A is chosen, then B must be chosen. In how many ways can selections be made? (3 Marks)

Ans. According to the question,

Total Number of Persons = 8
Number of People to be Selected = 6

It is given that, if A is chosen then, B must be chosen. Therefore, the following cases arise:

Case 1: When A is Chosen, B must be Chosen.

Number of Ways = Number of Ways of Selecting 4 persons from the Remaining 6 persons.

= ^8-2C_6-2

= ⁶C₄

= 15 Ways

Case 2: When A is not Chosen

Number of Ways = Number of Ways of Selecting 6 persons from Remaining 7 persons.

= ⁷C₆

Total Number of Ways to make Selection = 15 + 7 = 22 Ways

Ques. How many automobile license plates can be made if each plate contains two different letters followed by three different digits? (3 Marks)

Ans. There are 26 English alphabets and 10 digits (0 to 9).

It is given that each plate contains two different letters followed by three different digits.

Arrangement of 26 letters, taken 2 at a time = ¹⁰C₂ = 26 x 25 = 650
Arrangement of 10 digits taken three at a time = ¹⁰P₃ = 10 x 9 x 8 = 720

Total Number of License Plate = 650 x 720 = 468000

Ques. A bag contains 5 black and 6 red balls. Determine the number of ways in which 2 black and 3 red balls can be selected from the lot. (3 Marks)

Ans. The bag contains 5 black and 6 red balls.

2 black balls can be selected from 5 black balls in ⁵C₂ ways.
3 red balls can be selected from 6 red balls in ⁶C₃ ways.

Total number of ways in which 2 black and 3 red balls can be selected = ⁵C₂ x ⁶C₃

= 10 x 20 = 200

Thus, there are 200 ways in which 2 black and 3 red balls can be selected from the lot.

Ques. There are 10 lamps in a hall in which each one of them can be switched on independently. Find the number of ways in which the hall can be illuminated. (3 Marks)

Ans. There are 10 lamps in the hall. The hall can be illuminated if at least one lamp is switched.

Total number of Ways = ¹⁰C₁+ ¹⁰C₂ + ¹⁰C₃… + ¹⁰C₁₀

= 2¹⁰– 1

= 1024 - 1

= 1023

Thus, the hall can be illuminated in 1023 ways.

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Permutations and Combinations Formulas: Derivation, Example