CAT 2025 Lesson : Permutations & Combinations - Basics of Combinations

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: $6$

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: $56$