MHB In how many ways can we create such a committee?

[mathmari]

Hey!

From $16$ men and $12$ women we want to create a committee of $10$ people, with at least $3$ men and at least $5$ women. In how many ways can we create such a committee (men and women are discrete) ?

For that do we have to calculate the number of ways without restrictions minus the number of ways that the number of men is less than $3$ and the number of women less than $5$?

(Wondering)

 

[HOI]

1) Choose a man to be on the committee. There are 16 ways we can do that. There are then 15 men and 12 women left.
2) Choose another man to be on the committee. There are 15 ways we can do that. There are then 14 men and 12 women left.
3) Choose another man to be on the committee. There are 14 ways we can do that. There ae then 13 men and 12 women left.
There are 16(15)(14)= \frac{16!}{13!} ways to choose the three men to be on the committe in that order. Since order is not important, divide by 3! to get the number of ways to choose the three men: the binomial coefficient \begin{pmatrix}16 \\ 3 \end{pmatrix}= \frac{16!}{13!3!}.

(4) There are 12 ways to choose a woman to be on the committee. There are then 13 men and 11 women left.
(5) There are 11 ways to choose another woman to be on the committee. There are then 13 men and 10 women left.
(6) There are 10 ways to choose another woman to be on the committee. There are then 13 men and 9 women left.
(7) There are 9 ways to choose another woman to be on the committee. There are then 13 men and 8 women left.
(8) There are 8 ways to choose another woman to be on the committee. There are then 13 men and 7 women left.
There are 12(11)(10)(9)(8)= \frac{12!}{7!} ways to choose 5 women in that order. Since order is not important divide by 5! to get the number of ways to choose 5 women in any order: the binomial coefficient \begin{pmatrix}12 \\ 5\end{pmatrix}= \frac{12!}{7! 5!}. Since we want a committee of 10 people and so far we have only the minimal 3 men and 5 women adding to 8 people, we need to choose 2 more from all 13+ 7= 20 people left. There are (20)(19)/2= \frac{20!}{18!2!}= \begin{pmatrix}20 \\ 2 \end{pmatrix} ways to do that.

There are \begin{pmatrix}16 \\ 3\end{pmatrix}\begin{pmatrix}12 \\ 5\end{pmatrix}\begin{pmatrix}20\\ 2\end{pmatrix}= \frac{16!}{3!13!}\frac{12!}{7!5!<br /> }\frac{20!}{18!2!} ways to choose such a committee.

It seems strange that there are more men than women in the group but the minimum number of women on the committee is greater than the minimum number of men. A bit sexist isn't it?

 

[Opalg]

1) Choose a man to be on the committee. There are 16 ways we can do that. There are then 15 men and 12 women left.
2) Choose another man to be on the committee. There are 15 ways we can do that. There are then 14 men and 12 women left.
3) Choose another man to be on the committee. There are 14 ways we can do that. There ae then 13 men and 12 women left.
There are 16(15)(14)= \frac{16!}{13!} ways to choose the three men to be on the committe in that order. Since order is not important, divide by 3! to get the number of ways to choose the three men: the binomial coefficient \begin{pmatrix}16 \\ 3 \end{pmatrix}= \frac{16!}{13!3!}.

(4) There are 12 ways to choose a woman to be on the committee. There are then 13 men and 11 women left.
(5) There are 11 ways to choose another woman to be on the committee. There are then 13 men and 10 women left.
(6) There are 10 ways to choose another woman to be on the committee. There are then 13 men and 9 women left.
(7) There are 9 ways to choose another woman to be on the committee. There are then 13 men and 8 women left.
(8) There are 8 ways to choose another woman to be on the committee. There are then 13 men and 7 women left.
There are 12(11)(10)(9)(8)= \frac{12!}{7!} ways to choose 5 women in that order. Since order is not important divide by 5! to get the number of ways to choose 5 women in any order: the binomial coefficient \begin{pmatrix}12 \\ 5\end{pmatrix}= \frac{12!}{7! 5!}. Since we want a committee of 10 people and so far we have only the minimal 3 men and 5 women adding to 8 people, we need to choose 2 more from all 13+ 7= 20 people left. There are (20)(19)/2= \frac{20!}{18!2!}= \begin{pmatrix}20 \\ 2 \end{pmatrix} ways to do that.

There are \begin{pmatrix}16 \\ 3\end{pmatrix}\begin{pmatrix}12 \\ 5\end{pmatrix}\begin{pmatrix}20\\ 2\end{pmatrix}= \frac{16!}{3!13!}\frac{12!}{7!5!<br /> }\frac{20!}{18!2!} ways to choose such a committee.

It seems strange that there are more men than women in the group but the minimum number of women on the committee is greater than the minimum number of men. A bit sexist isn't it?

I think that there is some double counting in that approach, because of the way that the eight minimal members are selected first, followed by the two "top-up" members. For example, take the case of a committee consisting of four chosen men and six chosen women. if we select three of these four men as the "minimal" members then there are 4 ways of choosing those three out of four. Similarly there are 6 ways of choosing the "minimal" women. So this particular committee has been counted 24 times in Country Boy's enumeration.

There are three possible (mutually disjoint) ways to make up the committee: (i) 3 men and 7 women, (ii) 4 men and 6 women, (iii) 5 men and 5 women. The total number of ways to choose the committee is then $${16\choose 3}{12\choose 7} + {16\choose 4}{12\choose 6} + {16\choose 5}{12\choose 5} = 5\,584\,656.$$ That compares with Country Boy's total of $84\,268\,800.$