Question

Out of 6 men and 3 women, a committee of 4 is to be formed. In how many different ways it can be done if the committee must have:
At least 2 men ?
3 men and 1 woman ?​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that,

↝ Number of men = 6

↝ Number of women = 3

↝ Out of 6 men and 3 women, a committee of 4 is to be formed.

Part (a)

Out of 6 men and 3 women, a committee of 4 is to be formed must have atleast 2 men.

↝ So, Following cases arises

$\begin{gathered}\boxed{\begin{array}{c|c} \bf Men & \bf Women \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf 2 & \sf 2 \\ \\ \sf 3 & \sf 1\\ \\ \sf 4 & \sf 0 \end{array}} \\ \end{gathered}$

So, number of ways are

$\rm \:  =  \:^6C_{2} \times \: ^3C_{2} + \: ^6C_{3} \times \: ^3C_{1} + \: ^6C_{4} \times \: ^3C_{0}$

$\rm \:  =  \:\dfrac{6 \times 5}{2 \times 1} \times \dfrac{3}{1} + \dfrac{6 \times 5 \times 4}{3 \times 2 \times 1} \times \dfrac{3}{1} + \dfrac{6 \times 5}{2 \times 1} \times 1$

$\rm \:  =  \:45 + 60 + 15$

$\rm \:  =  \:120$

Part (b)

↝ Out of 6 men and 3 women, a committee of 4 is to be formed must have 3 men and 1 woman.

↝ So, number of ways are

$\rm :\longmapsto\:^6C_{3} \: \times \: ^3C_{1}$

$\rm \:  =  \:\dfrac{6 \times 5 \times 4}{3 \times 2 \times 1} \times 3$

$\rm \:  =  \:60$

Basic Concept Used :-

Number of ways of choosing r objects, out of n distinct objects is

$\rm :\longmapsto\:\boxed{ \tt{ \: Number \: of \: ways \: = \: ^nC_{r} \: }}$

Additional Information :

$\rm :\longmapsto\:\boxed{ \tt{ \: ^nC_{r} \: = \: ^nC_{n - r} \: }}$

$\rm :\longmapsto\:\boxed{ \tt{ \: ^nC_{0} \: = \: ^nC_{n} = 1\: }}$

$\rm :\longmapsto\:\boxed{ \tt{ \: ^nC_{1} \: = \: ^nC_{n - 1} = n\: }}$

$\rm :\longmapsto\:\boxed{ \tt{ \: ^nC_{r} \: = \frac{n}{r} \: ^{n - 1}C_{r - 1} \: }}$

$\rm :\longmapsto\:\boxed{ \tt{ \: \frac{^nC_{r}}{^nC_{r - 1}} = \frac{n - r + 1}{r} \: }}$

$\rm :\longmapsto\:\boxed{ \tt{ \: ^nC_{r} \: + \: ^nC_{r - 1} \: = \: ^{n + 1}C_{r} \: }}$