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Modern Maths

Permutations & Combinations

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CAT 2025 Lesson : Permutations & Combinations - Basics of Combinations

1.3 Basics of Combination

Combination relates to the selection of elements only. The order of arrangement does not matter.

The number of ways in which r elements can be selected from n distinct elements is

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

The following example explains the logic behind this formula.

Example 3

In how many ways can

2

students be selected from a group of

4

students?

Solution

This is similar to Example

\bm{1}

. However, the order in which the students are selected does not matter here.

Let the

4

students be A, B, C and D.

Number of ways

4

students can be arranged

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

Below are the

24

different ways in which

4

students can be arranged.

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

Like in the case of permutation, the arrangement of unselected students (

3^{\mathrm{rd}}

and

4^{\mathrm{th}}

in each arrangement) does not matter. These

2

unselected students can be arranged in

\bold{2!}

ways.

\therefore

Number of ways

4

students can be arranged in

2

places

= \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

However, in case of combinations, the arrangement of selected students (

1^{\mathrm{st}}

and

2^{\mathrm{nd}}

in each arrangement) does not matter. For instance, A B and B A are the same. The order in which they are selected does not make a difference. These

\bm{2}

selected elements can be arranged in

\bm{2!}

ways.

\therefore

Number of ways

2

students can be selected from

4 = \dfrac{12}{2!} = 6

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

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

\bold{r}

elements need to be selected, then there are

\bm{(n - r)}

elements that are not selected. The arrangement of these selected elements and the unselected elements do not matter in

\bm{n!}

. The selected elements and unselected elements can be arranged in

\bm{r!}

and

\bm{(n – r)!}

ways respectively.

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

In this question,

2

students are to be selected from

4

students.

Number of ways

= \space ^{4}C_2 = \dfrac{4!}{2!(4 - 2)!} = \dfrac{24}{2 \times 2} = 6

Answer:

6

So, the

^{n}C_r

formula is the number of arrangements of

n

elements, i.e.

\bm{n!}

, divided by the number of arrangements of

r

selected elements, i.e.

\bm{r!}

, and the number of arrangements of

(n - r)

unselected elements, i.e.

\bm{(n - r)!}

Example 4

In how many ways can

5

students be selected from a group of

8

students?

Solution

Total number of students =

\bm{n} = 8

Number of students to be selected

= \bm{r} = 5

Total number of combinations/selections

= \space ^{n}C_r = \space ^{8}C_5 = \dfrac{8!}{5!(8 - 5)!} = \dfrac{8!}{5! \times 3!} = \dfrac{8 \times 7 \times 6}{3!} = 56

Answer:

56

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