Question

Out of 8 men and 10 women a committee consisting of 6 men and 5 women is to be made. How many committees will be made if a particular man A refuses to be a member of the committee in which a particular lady B is there?

Answer 1

For solving this question, we will first find the total number of ways of forming a committee with 6 men and 5 women out of 8 men and 10 women. Then we will find the number of ways of forming a committee when both A and B are there in the committee. Subtracting these, we will get the required number of ways. We will use the formula of combination to find the number of ways. Number of ways of selecting r items out of n item is given by ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$.

Complete step by step answer:
Here we are given a number of men as 8 and a number of women as 10. We need to select a committee having 6 men and 5 women but out of these 8 men, a man A refuses to be in committee if a woman B is in the committee. So let us find required ways of selecting such a committee.
First let us find ways of selecting a committee without any restriction. We have 8 women out of which 6 are to be selected. We know number of ways to select r items out of n is given by ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$.
So here we have, number of ways of selecting men $\Rightarrow {}^{8}{{C}_{6}}$.
Similarly, for selecting 5 women out of 10 we have a number of ways of selecting women $\Rightarrow {}^{10}{{C}_{5}}$.
Total ways of selecting committee becomes $\Rightarrow {}^{8}{{C}_{6}}\times {}^{10}{{C}_{5}}\cdots \cdots \cdots \left( 1 \right)$.
Now let us find the number of ways to form a committee when both A and B are there (which we do not require) so that we can subtract it from total ways.
If A is already selected so we have 7 men left and 5 positions left. Hence the number of ways of selecting 5 positions for men $\Rightarrow {}^{7}{{C}_{5}}$.
Similarly if B is selected, we have 9 women left and 4 positions left. Hence the number of ways of selecting 4 positions for women $\Rightarrow {}^{9}{{C}_{4}}$.
Hence number of ways of selecting a committee where both A and B are there is $\Rightarrow {}^{7}{{C}_{5}}\times {}^{9}{{C}_{4}}\cdots \cdots \cdots \left( 2 \right)$.
Now let us subtract (2) from (1) to get required ways, we get,
Required ways $\Rightarrow {}^{8}{{C}_{6}}\times {}^{10}{{C}_{5}}-{}^{7}{{C}_{5}}\times {}^{9}{{C}_{4}}$.
Using ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$ we have,

& \Rightarrow \dfrac{8!}{6!2!}\times \dfrac{10!}{5!5!}-\dfrac{7!}{5!2!}\times \dfrac{9!}{4!5!} \\\ & \Rightarrow \dfrac{8\times 7\times 6!}{6!\times 2}\times \dfrac{10\times 9\times 8\times 7\times 6\times 5!}{5!\times 5\times 4\times 3\times 2}-\dfrac{7\times 6\times 5!}{5!\times 2}\times \dfrac{9\times 8\times 7\times 6\times 5!}{5!\times \times 4\times 3\times 2} \\\ \end{aligned}$$. Cancelling the terms we get, $$\begin{aligned} & \Rightarrow 4\times 7\times 9\times 2\times 7\times 2-7\times 3\times 3\times 7\times 6 \\\ & \Rightarrow 7056-2646 \\\ & \Rightarrow 4410 \\\ \end{aligned}$$ **Hence the required number of ways of selecting a committee where A and B are not together is 4410.** **Note:** Students should be careful while calculating ${}^{n}{{C}_{r}}$ as complex calculations are involved. Students can also find ways using the following method. Finding ways of selecting committees when both A and B are not together, finding ways when only A is present and finding ways when only B is present. Adding all can also give a solution.