Q. 1.12 - A First Course in Probability

Question

A committee of $6$ people is to be chosen from a group consisting of $7$ men and $8$ women. If the committee must consist of at least $3$ women and at least $2$ men, how many different committees are possible?

Step-by-Step Solution

Answer

The possible no. of committees are $3430$.

Step 1. Given information.

Total no. of men $=7$

Total no. of women $=8$

Total no. of members in committee $=6$

Minimum no. of women in committee $=3$

Minimum no. of men in committee $=2$

There can be two ways of choosing the committee, either $3$ men and $3$ women or $2$ men and $4$ women.

Step 2. Find the no. of ways of choosing 3 men and 3 women for committee.

$3$ men out of $7$ can be chosen in $\left(\begin{array}{c}7\\ 3\end{array}\right)\text{ways}=\frac{7!}{3!4!}=35$

$3$ women out of $8$ can be chosen in $\left(\begin{array}{c}8\\ 3\end{array}\right)\text{ways}=\frac{8!}{3!5!}=56$

Therefore, the no. of ways of choosing a committee of $3$ men and $3$ women $=35\times 56=1960$

Step 3. Find the no. of ways of choosing 2 men and 4 women for committee.

$2$ out of $7$ men can be chosen in $\left(\begin{array}{c}7\\ 2\end{array}\right)\text{ways}=\frac{7!}{2!5!}=21$

$4$ out of $8$ women can be chosen in $\left(\begin{array}{c}8\\ 4\end{array}\right)\text{ways}=\frac{8!}{4!4!}=70$

Therefore, the no. of ways of choosing a committee of $2$ men and $4$ women $=21\times 70=1470$

Step 4. Find the possible no. of committees.

Therefore, the possible no. of committees are $=1960+1470=3430$.