Question

From a group of 8 women and 6 men, a committee of 3 men and 3 women is to be formed. In how many ways can the committee be formed if one man and one woman refused to serve together?
A. 722
B. 784
C. 910
D. 896

Answer 1

Hint: We will analyze the question and apply the given condition, we are required to make a committee of 3 men and 3 women and then we will form three conditions under it one where a man is selected, second where the woman is selected and third when no one is selected and we will find the ways by applying: ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$ and finally summing them up to get the answer.

Complete step-by-step solution
Now, we are given that we have 8 women and 6 men and we need to form a committee of 3 men and 3 women.
Now, let the man who refuses to work together be M and the woman who refuses to work together be W. Now, we have to find out the number of ways in which the committees need to be formed if M and W refuse to serve together. So, we will see the conditions in which they can be done:
Condition I: If M is selected and W is not:
If M is selected then we need to select 2 more men from the remaining 5 men and since W is not selected we will have 7 women and we have to select 3 out of them, so we will use the combination formula that is: ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$, where n is the total number of object and r is the selected number of object.
Therefore, the total number of ways in which 3 men and 3 women selected in this condition is as follows:
$\Rightarrow {}^{7}{{C}_{3}}\times {}^{5}{{C}_{2}}=\dfrac{7!}{3!\left( 7-3 \right)!}\times \dfrac{5!}{2!\left( 5-2 \right)!}=35\times 10=350$
Condition II: If W is selected and M is not:
If W is selected then we need to select 2 more women from the remaining 7 women and since M is not selected we will have 5 men and we have to select 3 out of them, so we will again use the combination formula that is: ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$
Therefore, the total number of ways in which 3 men and 3 women selected in this condition is as follows:
$\Rightarrow {}^{7}{{C}_{2}}\times {}^{5}{{C}_{3}}=\dfrac{7!}{2!\left( 7-2 \right)!}\times \dfrac{5!}{3!\left( 5-3 \right)!}=21\times 10=210$
Condition II: If both W and M are not selected:
If W is not selected then we need to select 3 women from the remaining 7 women and since M is not selected we will have 5 men and we have to select 3 out of them, so we will again use the combination formula that is: ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$
Therefore, the total number of ways in which 3 men and 3 women selected in this condition is as follows:
$\Rightarrow {}^{7}{{C}_{3}}\times {}^{5}{{C}_{3}}=\dfrac{7!}{3!\left( 7-3 \right)!}\times \dfrac{5!}{3!\left( 5-3 \right)!}=35\times 10=350$
Now, we will add all three to get the total number of ways: $350+210+350=910$
Hence, the correct option is C.

Note: A common mistake in these cases is while solving the given factorials, as we know that the calculations can be a bit tricky, we can use some shortcuts to solve the factorials easily. For example:
${}^{n}{{C}_{r}}={}^{n}{{C}_{n-r}}$ , like in above question we can see that: ${}^{5}{{C}_{3}}={}^{5}{{C}_{5-3}}={}^{5}{{C}_{2}}$ and we can also apply: $n!=n\times \left( n-1 \right)!$ . These can come handy while solving the questions.