Q. 1.19 - A First Course in Probability

Question

From a group of $8$ women and $6$ men, a committee consisting of $3$ men and $3$ women is to be formed. How many

different committees are possible if

(a) $2$ of the men refuse to serve together?

(b) $2$ of the women refuse to serve together?

(c) $1$ man and $1$ woman refuse to serve together?

Step-by-Step Solution

Answer

(a) The possible number of different committees are $896$.

(b) The possible number of different committees are $1000$.

(c) The possible number of different committees are $910$.

Part (a) Step 1. Given information.

Total no. of men $=6$.

No. of men to be selected $=3$.

Total no. of women $=8$.

No. of women to be selected $=3$

Part (a) Step 2. Find the number of different committees if 2 of the men refuse to serve together.

The number of ways to form committee, when the two men who refused to work together are excluded

$\left(\begin{array}{c}8\\ 3\end{array}\right)\left(\begin{array}{c}4\\ 3\end{array}\right)$ $=\frac{8!}{3!5!}\times \frac{4!}{3!1!}=\frac{8\times 7\times 6\times 5!}{3\times 2\times 1\times 5!}\times \frac{4\times 3!}{3!}=224$

The two men who refused to work together can be any of the $6$ men, therefore the possible different combinations are

$\left(\begin{array}{c}8\\ 3\end{array}\right)\left(\begin{array}{c}4\\ 2\end{array}\right)\left(\begin{array}{c}2\\ 1\end{array}\right)=\frac{8!}{3!5!}\times \frac{4!}{2!2!}\times \frac{2!}{1!1!}=\frac{8\times 7\times 6\times 5!}{3\times 2\times 1\times 5!}\times \frac{4\times 3\times 2!}{2\times 2!}\times 2\times 1=672$

Therefore, the possible number of different committees are $=224+672=896$.

Part (b) Step 3. Find the number of different committees if 2 of the women refuse to serve together.

The number of ways to form committee, when the two men who refused to work together are excluded

$\left(\begin{array}{c}6\\ 3\end{array}\right)\left(\begin{array}{c}6\\ 3\end{array}\right)=\frac{6!}{3!3!}\times \frac{6!}{3!3!}=\frac{6\times 5\times 4\times 3!}{3\times 2\times 1\times 3!}\times \frac{6\times 5\times 4\times 3!}{3\times 2\times 1\times 3!}=400$

The two women who refused to work together can be any of the $8$ women, therefore the possible different combinations are

$\left(\begin{array}{c}6\\ 2\end{array}\right)\left(\begin{array}{c}6\\ 3\end{array}\right)\left(\begin{array}{c}2\\ 1\end{array}\right)=\frac{6!}{2!4!}\times \frac{6!}{3!3!}=\frac{6\times 5\times 4!}{2\times 1\times 4!}\times \frac{6\times 5\times 4\times 3!}{3\times 2\times 1\times 3!}\times 2=600$

Therefore, the possible number of different committees are $=400+600=1000$

Part (c) Step 4. Find the number of different committees if 1 man and 1 woman refuse to serve together.

The number of ways to form committee, when the man and woman who refused to work together are excluded

$\left(\begin{array}{c}7\\ 3\end{array}\right)\left(\begin{array}{c}5\\ 3\end{array}\right)=\frac{7!}{3!4!}\times \frac{5!}{3!2!}=\frac{7\times 6\times 5\times 4!}{3\times 2\times 1\times 4!}\times \frac{5\times 4\times 3!}{2\times 1\times 3!}=350$

The number of ways to form committee, when the woman is included but the man is excluded

$\left(\begin{array}{c}7\\ 2\end{array}\right)\left(\begin{array}{c}5\\ 3\end{array}\right)\left(\begin{array}{c}1\\ 1\end{array}\right)=\frac{7!}{2!5!}\times \frac{5!}{3!2!}=\frac{7\times 6\times 5!}{2\times 1\times 5!}\times \frac{5\times 4\times 3!}{2\times 1\times 3!}\times 1=210$

The number of ways to form committee, when the man is included but the woman is excluded 

$\left(\begin{array}{c}7\\ 3\end{array}\right)\left(\begin{array}{c}5\\ 2\end{array}\right)\left(\begin{array}{c}1\\ 1\end{array}\right)=\frac{7!}{3!4!}\times \frac{5!}{2!3!}=\frac{7\times 6\times 5\times 4!}{3\times 2\times 1\times 4!}\times \frac{5\times 4\times 3!}{2\times 1\times 3!}\times 1=350$

Therefore, the possible number of different committees are $=350+210+350=910$