how many ways can the committee be formed if one women $W_1$ refuses to be in the committee if another particular women $W_2​$ be included?

Question

There are $7$ men and $5$ women who are eligible for being included in a committee of $4$ men and $3 $ women. In how many ways can the committee be formed if one women $W_1$ refuses to be in the committee if another particular women $W_2​$ be included?

My solution

• Case 1 $W_1$ included. Therefore $W_2$ will not join.($\because$ if one women $W_1$ refuses to be in the committee if another particular women $W_2​$ be included)

$\therefore$ Total number of ways we can selected the committee is $7 C_4 \times 3C_2$($\because$ we select $4$ men out of $7$ men and If $W_1$ already selected and $W_2$ is not included, we have only 5-2 women there. We already selected $W_1$, so we need to select two more women to form the committee.)

Case 2 $W_2$ included. Therefore $W_1$ will not join.($\because$ if one women $W_1$ refuses to be in the committee if another particular women $W_2$ be included)

$\therefore$ Total number of ways we can selected the committee is $7 C_4 \times 3C_2$($\because$ we select $4$ men out of $7$ men and If $W_2$ already selected and $W_1$ is not included, we have only $5-2=3$ women there. We already selected $W_1$, so we need to select two more women to form the committee.)

Is my logic correct? Answer given was different. Answer given was 280. But I got 210, by adding the two cases.

Answer 1

You forgot to include the case when none of them are included in the committee. Note that this still satisfies the condition of the problem since if none of them are in the committee, then they do not belong to the same committee.

In this case, you have to select $4$ of the $7$ men and all of the remaining $3$ women, so there are $\binom{7}{4}\binom{3}{3} = 35$ ways to form the committee.

In total, there are $210 + 35 = 245$ ways to form the committee.

Another way of solving it without doing the casework is to count the number of ways that a committee can be formed (without having to restrict $W_1$ or $W_2$ from joining the committee) and subtract it by the number of ways to form a committee with both $W_1$ and $W_2$ included. You will get the same no. of ways: $$\binom{7}{4}\binom{5}{3} - \binom{7}{4}\binom{3}{1} = 245$$