Question

A committee of $6$ is chosen from $10$ men and $7$ women so as to contain at least $3$ men and $2$ women. In how many ways can this be done if two particular women refuse to serve on the same committee?
$\text{A) 7800}$
$\text{B) 8610}$
$\text{C) 810}$
$\text{D) 7700}$

Answer 1

Hint: In this question we have to find the number of ways we can select a committee of $6$ members given that there are at least $3$ men and $2$women out of the total of $10$ men and $7$ women which are available. We will solve the question by finding all the ways in which $6$ members can be selected from which we will subtract the number of ways in which the two women are together to get the required solution.

Complete step by step solution:
We know that there should be at least $3$ men and $2$women out of the total of $10$ men and $7$ women therefore the possible cases are:
$3$ men and $3$ women and
$4$ men and $2$ women.
Now the number of ways $3$ men and $3$ women can be selected out of the total of $10$ men and $7$ women can be represented using the combination formula as:
$\Rightarrow {}^{10}{{C}_{3}}{{\times }^{7}}{{C}_{3}}$
And the number of ways $4$ men and $2$ women can be selected out of the total of $10$ men and $7$ women can be represented using the combination formula as:
$\Rightarrow {}^{10}{{C}_{4}}{{\times }^{7}}{{C}_{2}}$
Now to get the total number of cases we will add both the cases. On adding, we get:
$\Rightarrow {}^{10}{{C}_{3}}{{\times }^{7}}{{C}_{3}}+{}^{10}{{C}_{4}}{{\times }^{7}}{{C}_{2}}$, this represents the total number of ways.
Now consider the cases in which the two women are together. If the two women are together in the committee there is space for only $4$ other members.
The division can be as:
 $3$ men and $1$ other woman out of the $5$ remaining women and
 $4$ men and $0$ other women given $2$ women are already present.
And the number of ways $3$ men and $1$ women can be selected out of the total of $10$ men and $5$ women can be represented using the combination formula as:
${{\Rightarrow }^{10}}{{C}_{3}}{{\times }^{5}}{{C}_{1}}$
And the number of ways $4$ men and $0$ women can be selected out of the total of $10$ men and $5$ women can be represented using the combination formula as:
${{\Rightarrow }^{10}}{{C}_{4}}{{\times }^{5}}{{C}_{0}}$
Now to get the total number of cases we will add both the cases. On adding, we get:
$\Rightarrow {}^{10}{{C}_{3}}{{\times }^{5}}{{C}_{1}}+{}^{10}{{C}_{4}}{{\times }^{5}}{{C}_{0}}$.
Now to get the required number of ways we will subtract the unwanted number of ways from the total number of ways. On subtracting, we get:
$\Rightarrow {}^{10}{{C}_{3}}{{\times }^{7}}{{C}_{3}}+{}^{10}{{C}_{4}}{{\times }^{7}}{{C}_{2}}-{}^{10}{{C}_{3}}{{\times }^{5}}{{C}_{1}}+{}^{10}{{C}_{4}}{{\times }^{5}}{{C}_{0}}$
On simplifying the values using a scientific calculator, we get:
$\Rightarrow 8610-810$
On simplifying, we get:
$ 7800$, which is the required answer. Therefore the correct option is $\left( A \right)$.

Note: It is to be remembered that in this question we have used the combination formula and not the permutation formula since we have to find the total number of possible ways and the order does not matter in this question. The formula for combination should be remembered which is
$Combinations={}^{n}{{C}_{r}}=n!(n-r)!$