Question

The number of ways of choosing a committee of two women and three men from five women and six men, if Mr. A refuses to serve on the committee if Mr. B is a member and Mr. B can only serve, if Miss C is the member of the committee is
(a) 60
(b) 84
(c) 124
(d) none of these

Answer 1

We will divide the question into three cases and find the number of ways in each case and then we will add all the ways to get the answer. First case is where miss C is a member then Mr. B is also a member and then Mr. A cannot be a member. Second case is where Miss C is a member but Mr. B is not a member and this implies Mr. A can or cannot be a member. And the third case is where miss C is not a member then Mr. B is also not a member and this implies Mr. A can or cannot be a member.

Complete step-by-step answer:
It is mentioned in the question that the committee should have two women and three men. And the total number of women is 5 and the total number of men is 6.
So let’s take the first case where miss C is a member then Mr. B is also a member and this implies Mr. A cannot be a member. Hence we have 1 woman and 1 man. So now we have to select one woman from the remaining 4 women as miss C is already a member. Similarly we have to select 2 men from the remaining 4 men as Mr. B is already a member and Mr. A cannot be a member. So the number of ways of choosing a committee $={}^{4}{{C}_{1}}\times {}^{4}{{C}_{2}}......(1)$. Now applying the formula ${}^{n}{{C}_{r}}=\dfrac{n!}{r!(n-r)!}$ in equation (1), we get,

& \Rightarrow \dfrac{4!}{1!(4-1)!}\times \dfrac{4!}{2!(4-2)!} \\\ & \Rightarrow \dfrac{4!}{1!3!}\times \dfrac{4!}{2!2!} \\\ & \Rightarrow \dfrac{4\times 3!}{1!3!}\times \dfrac{4\times 3\times 2}{2\times 2}=4\times 6=24 \\\ \end{aligned}$$ So the number of ways of choosing the committee in this case is 24. Now let’s take the second case where miss C is a member but Mr. B is not a member and this implies Mr. A can or cannot be a member. Hence we have 1 woman and 0 men. So now we have to select one woman from the remaining 4 women as miss C is already a member. Similarly we have to select 3 men from 5 men as there are 0 men and B has denied to be a member now. So the number of ways of choosing a committee $$={}^{4}{{C}_{1}}\times {}^{5}{{C}_{3}}......(2)$$. Now applying the formula $${}^{n}{{C}_{r}}=\dfrac{n!}{r!(n-r)!}$$ in equation (2), we get, $$\begin{aligned} & \Rightarrow \dfrac{4!}{1!(4-1)!}\times \dfrac{5!}{3!(5-3)!} \\\ & \Rightarrow \dfrac{4!}{1!3!}\times \dfrac{5!}{3!2!} \\\ & \Rightarrow \dfrac{4\times 3!}{1!3!}\times \dfrac{5\times 4\times 3!}{3!\times 2}=4\times 10=40 \\\ \end{aligned}$$ So the number of ways of choosing the committee in this case is 40. Now let’s take the third case where miss C is not a member then Mr. B is also not a member and this implies Mr. A can or cannot be a member. Hence we have 0 women and 0 men. So now we have to select 2 women from the remaining 4 women as miss C has denied to be a member. Similarly we have to select 3 men from 5 men as there are 0 men and B cannot be a member now. So the number of ways of choosing a committee $$={}^{4}{{C}_{2}}\times {}^{5}{{C}_{3}}......(3)$$. Now applying the formula $${}^{n}{{C}_{r}}=\dfrac{n!}{r!(n-r)!}$$ in equation (3), we get, $$\begin{aligned} & \Rightarrow \dfrac{4!}{2!(4-2)!}\times \dfrac{5!}{3!(5-3)!} \\\ & \Rightarrow \dfrac{4!}{2!2!}\times \dfrac{5!}{3!2!} \\\ & \Rightarrow \dfrac{4\times 3\times 2!}{2!2!}\times \dfrac{5\times 4\times 3!}{3!\times 2}=6\times 10=60 \\\ \end{aligned}$$ So the number of ways of choosing the committee in this case is 60. So now the total ways of choosing the committee $$=24+40+60=124$$ **Hence the correct answer is option (c).** **Note:** Remembering the formula $${}^{n}{{C}_{r}}=\dfrac{n!}{r!(n-r)!}$$ and the concept of combinations is the key here. A combination is a way to order or arrange a set or number of things uniquely. We in a hurry can make a mistake in solving equation (1), equation (2) and equation (3) and hence we need to be careful while doing these steps.