Question

In how many ways can a committee of $4$ be selected from $6$ men and $8$ women if the committee must contain at most $2$ women?

Answer 1

We will find the number of ways the committee can be selected from $6$ men and $8$ women if the committee must contain at most $2$ women by considering the committee with $0$ women, $1$ woman and $2$ women.

Complete step by step solution:
We are asked to find the number of ways a committee of $4$ can be formed. We have a total of $6$ men and a total of $8$ women.
In any case, we can select a maximum of $2$ women for the committee.
So, we can form the committee with no woman. That is, we can select $4$ men.
Or we can form the committee with $1$ woman. So, there will be $3$ men and $1$ woman in the committee.
Finally, the committee can be formed with $2$ women. So, the committee will have $2$ men and $2$ women.
Let us consider the committee with $0$ woman.
In that case, we will get the committee with $4$ men as we have already mentioned. Since there are $6$ men out of which we need to select $4,$ we will use a combination.
So, we will get ${}^{6}{{C}_{4}}=\dfrac{6\times 5\times 4\times 3}{1\times 2\times 3\times 4}=\dfrac{30}{2}=15.$
Now, we will find the number of ways the committee can be formed with $1$ women.
Since there are $8$ women and $6$ men, we can choose $1$ woman from them by ${}^{8}{{C}_{1}}=8$ and $3$ from them by ${}^{6}{{C}_{3}}=\dfrac{6\times 4\times 5}{1\times 2\times 3}=20.$ Therefore, we will get $8\times 20=160.$
Now, we will consider the case when there are $2$ women. We can choose $2$ women out of the $8$ women by ${}^{8}{{C}_{2}}=\dfrac{8\times 7}{1\times 2}=28$ and $2$ men out of the $6$ men by ${}^{6}{{C}_{2}}=\dfrac{6\times 5}{1\times 2}=15.$ Now, we will get $28\times 15=420.$
Now, we will add these numbers to get the number of ways the committee can be selected.

Hence the number of ways we can select the committee is $420+160+15=595.$

Note: We have used the combination to find the number of ways we can select the committee. Combination is used to determine the number possible arrangements of objects when the order does not matter.