Question

A committee of $6$ people is to be chosen from $5$ men and $8$ women. In how many ways can this be done
(i) If there are more women than men on the committee
(ii) If the committee consists of $3$ men and $3$ women but two particular men refuse to be on the committee together?
One particular committee consists of $5$ women and $1$ man.
(iii) In how many different ways can the committee members be arranged in a line if the man is not at either end?

Answer 1

For the first question, find the different cases where the number of women is greater than men and then find the combinations for it. For the second question, Subtract the number of ways two men refuse to sit together with the total number of possibilities. For the third part, find the combinations by placing two women on either side of the corners.

Complete step-by-step solution:
Given that there is a committee of $6$ people is to be chosen from $5$ men and $8$ women.
Solving the (i) part first,
Write down the case where there can more women than men which are,
There are six positions. \begin{array}{*{20}{c}} \\_&\\_&\\_&\\_&\\_&\\_ \end{array}
Now the cases and their combinations where the number of women should be greater than man is,
$\Rightarrow 0$ man $6$ women. The combination would be $^8{C_6}{ \times ^5}{C_0}$
$\Rightarrow 1$ man $5$ women. The combination would be $^8{C_5}{ \times ^5}{C_1}$
$\Rightarrow 2$ man $4$ women. The combination would be $^8{C_4}{ \times ^5}{C_2}$
Now add all of them to get the number of ways.
$\Rightarrow {(^8}{C_6}{ \times ^5}{C_0}) + {(^8}{C_5}{ \times ^5}{C_1}) + {(^8}{C_4}{ \times ^5}{C_2})$
Now solving the (ii) part.
Given that there are $3$ men and $3$ women
There are $6$ positions so, \begin{array}{*{20}{c}} \\_&\\_&\\_&\\_&\\_&\\_ \end{array}
The total possibilities are $^8{C_3}{ \times ^5}{C_3} = 56 \times 10 = 560$ ways
But given that two men refuse to serve together.
We can select two men from only two men and the other man can be selected from the remaining three men.
$= 2M + 1M + 3W$
Which can be combined as,
${ \Rightarrow ^2}{C_2}{ \times ^3}{C_1}{ \times ^8}{C_3}$
This is the case where the two men are the same team and then we selected one man from the remaining $(5 - 2) = 3$ and then selected three women from $8$ women.
$\Rightarrow 1 \times 3 \times 56 = 168$
Now we subtract it from the total number of ways.
${ \Rightarrow ^8}{C_3}{ \times ^5}{C_3}{ - ^2}{C_2}{ \times ^3}{C_1}{ \times ^8}{C_3}$
The ways in which two men are not together $560 - 168 = 392$ .
Now solving the (iii) part.
Place two women on either ends so there is no chance for a man on the ends.
\begin{array}{*{20}{c}} W&\\_&\\_&\\_&\\_&W; \end{array}Now, this can be arranged in,
${ \Rightarrow ^8}{C_5} \times 5!{ \times ^5}{C_1} \times 4$
First, we select $5$ women from $8$ women and they can be arranged in $5!\;$ ways. Now one man must be selected from $5$ men and then there are $4$ places left after the women occupy the ends so there are $4$ times the ways for the man to permute in any position.

$\therefore$ (i) ${(^8}{C_6}{ \times ^5}{C_0}) + {(^8}{C_5}{ \times ^5}{C_1}) + {(^8}{C_4}{ \times ^5}{C_2})$ ways
(ii) $^8{C_3}{ \times ^5}{C_3}{ - ^2}{C_2}{ \times ^3}{C_1}{ \times ^8}{C_3}$ ways
(iii) $^8{C_5} \times 5!{ \times ^5}{C_1} \times 4$ ways

Note: There is no need to solve the entire combination because it is much of a calculation. Write it in the combination form itself for the reader also to understand. Always
write down the cases separately and then write the permutations and combinations to it.