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Modern Maths

Permutations & Combinations

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CAT 2025 Lesson : Permutations & Combinations - Basics of Permutations

1. Introduction

Generally

2

3

questions are asked from Permutations & Combinations and Probability in CAT and other MBA entrance tests. These are perceived to be difficult chapters and are skipped very often. However, once you have a thorough understanding of the concepts, these questions take very little time to answer. Therefore, this is a scoring area which should not be skipped.

In this lesson, we will learn the basics of factorials

(!)

, permutation and combination, and look at the different possible types of questions. In simple terms,
Permutations is the number of ways in which items can be selected and arranged.
Combinations is the number of ways in which items can be selected only.

1.1 Factorial and its properties

Factorial is represented by an exclamation mark ('

!

'). '

\bm{n}!

' is read as '

\bm{n}

\bold{factorial}

\bm{n!}

is the product of all natural numbers less than or equal to

\bm{n}

.
⇒

n! = 1 \times 2... \times n

For example,

10! = 1 \times 2 \times 3 \times 4 \times 5 \times 6 \times 7 \times 8 \times 9 \times 10

Exception: The only exception to this definition is

0!

0

is not a natural number, however,

0!

is defined.

\bold{0!}

\bold{1.}

To enhance your speed, you should memorise the values of the following commonly used factorials.

Factorial	Value
$0!$	$1$
$1!$	$1$
$2!$	$2$
$3!$	$6$
$4!$	$24$
$5!$	$120$
$6!$	$720$
$7!$	$5,040$
$8!$	$40,320$

Note,

n! = n \times (n - 1)! = n \times (n - 1) \times (n - 2)! = ...

For instance,

13! = 13 \times 12! = 13 \times 12 \times 11! = ...

1.2 Basics of Permutation

Permutation refers to the arrangement of elements or things. The order of arrangement matters.

Note that

\bm{n}

elements can be arranged in

\bm{n!}

ways. However, if

\bm{r}

elements are to be selected from a set of

\bm{n}

distinct elements and then arranged, this can be done in

^{n}P_r = \dfrac{n!}{(n - r)!}

ways.

So, as the definition suggests, permutation includes selection and arrangement of elements. The following example explains the logic behind this formula.

Example 1

In how many ways can

4

students be seated on

2

chairs?

Solution

Let the

4

students be A, B, C and D. Let's begin by seating them in

4

chairs (instead of

2

).

In the

1^{\mathrm{st}}

chair, any of the

4

students can sit. In the

2^{\mathrm{nd}}

chair, any of the

3

remaining students can sit (as

1

student is already seated in the

1^{\mathrm{st}}

chair). Likewise,

2

students can sit on the

3^{\mathrm{rd}}

chair and

1

student in the

4^{\mathrm{th}}

chair. This concept is detailed in section 2.4 Using Blanks.

\therefore

the number of ways

4

students can sit in

4

chairs

= 4! = 4 \times 3\times 2 \times 1 = 24

Below are the

24

different ways in which

\bm{4}

students can be seated in

\bm{4}

chairs.

A B C D	B A C D	C A B D	D A B C
A B D C	B A D C	C A D B	D A C B
A C B D	B C A D	C B A D	D B A C
A C D B	B C D A	C B D A	D B C A
A D B C	B D A C	C D A B	D C A B
A D C B	B D C A	C D B A	D C B A

The question, however, requires us to arrange only

\bold{2}

of the

\bold{4}

students. So, the students seated in the

3^{\mathrm{rd}}

and

4^{\mathrm{th}}

chairs need not be arranged.

Let us take the case where A and B sit on the

1^{\mathrm{st}}

and

2^{\mathrm{nd}}

chairs respectively. This occurs twice in the above arrangements - A B C D and A B D C.

In this case, C and D, the unselected students, have been arranged as well. In every such case,

\bold{2}

unselected students are arranged in

\bold{2!}

ways. Therefore, this has to be divided from the total possibilities.

\therefore

Number of ways

4

students can sit in

2

chairs

= \dfrac{4!}{2!} = \dfrac{24}{2} = 12

A B	B A	C A	D A
A C	B C	C B	D B
A D	B D	C D	D C

Alternatively (Recommended Method)

If there are

\bm{n}

elements, then number of ways they can be arranged

= \bm{n}!

From these

\bm{n}

elements, if

\bm{r}

elements need to be selected and arranged, then there are

\bm{n - r}

elements that are not selected. The arrangement of these unselected elements in

\bm{n}!

does not matter. The unselected elements can be arranged in

\bm{n - r}!

ways.

\therefore \space ^{n}{P}_r = \dfrac{\bm{n}!}{\bm{n - r}!}

In this question,

2

students are to be selected from

4

students and then seated.

Number of ways

= \therefore \space ^{4}{P}_2 = \dfrac{\bm{4}!}{\bm{4 - 2}!} = \dfrac{24}{2} = 12

Answer: 12

So, the

^{n}{P}_r

formula is the number of arrangements of

\bm{n}

elements, i.e.

\bm{n!}

, divided by the number of arrangements of the unselected elements

(n - r)

, i.e.

\bm{(n - r)!}

Example 2

In how many ways can

4

students be seated on

4

chairs?

Solution

As all

4

students are to be arranged, and there are no unselected items, the total number of ways =

\bm{4!} = \bm{24}

Alternatively

Total number of students

= \bm{n} = 4

Number of students to be arranged

= \bm{r} = 4

Total number of permutations

= \space ^{n}{P}_r = \space ^{4}{P}_4 - \dfrac{4!}{(4 - 4)!} = \dfrac{4!}{0!} = 24

Answer: 24

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