Question

A committee of 5 men and 3 women is to be formed out of 7 men and 6 women. If two particular women are not to be together in the committee, the number of committees formed is,
A. 420
B. 5040
C. 336
D. 216

Answer 1

Hint:We will be using the concepts of permutation and combination to solve the problem. We will be forming different cases corresponding to the condition given in the question and sum them up to find the answer.

Complete step-by-step answer:
Now, we have been given 7 men and 6 women. We have to form a committee of 5 men and 3 women from this also it has been given that two particular women are not together. Now, we take the first case in which we take the combination of 5 men from 7 men and 3 women from 4 women excluding both the women which should not be together. So, we have,
Number of committees formed \[={}^{7}{{C}_{5}}\times {}^{4}{{C}_{3}}\]
Now, in the second case we again will take a combination of men from 7 men but for women we take 1 of the women which can’t be together and 2 from the remaining 4 women.
Number of committees formed \[={}^{7}{{C}_{5}}\times {}^{2}{{C}_{1}}\times {}^{4}{{C}_{2}}\]
So, the total possible number of committee formed \[={}^{7}{{C}_{5}}\times {}^{4}{{C}_{3}}+{}^{7}{{C}_{5}}\times {}^{2}{{C}_{1}}\times {}^{4}{{C}_{2}}\]
Now, we know that \[{}^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!\times r!}\].
So, we have,
$\begin{align}
  & =\dfrac{7!}{5!\times 2!}\times \dfrac{4!}{3!\times 1!}+\dfrac{7!}{5!\times 2!}\times \dfrac{2\times 4!}{2!\times 2!} \\
 & =\dfrac{6\times 7}{2}\times 4+\dfrac{6\times 7}{2}\times 2\times 6 \\
 & =21\times 4+21\times 2\times 6 \\
 & =84+42\times 6 \\
 & =84+252 \\
 & =336 \\
\end{align}$
So, we have the number of committees formed is 336.
Hence, the correct option is (C).

Note: To solve these type of questions it is important to note how we break the question into cases and find the answer by solving for different cases and then adding the answer of different cases to get the final answer.