# A club with 32 women and 10 men needs to form a committee of size 5

I want to make sure that I'm thinking of this correctly.

A club with 32 women and 10 men needs to form a committee of size 5:

(a) How many committees are possible? - I said $$\begin{pmatrix} 42 \\ 5 \\ \end{pmatrix}$$

(b) How many committees are possible if a committee must have 3 women and 2 men? - $$\begin{pmatrix} 32 \\ 3 \\ \end{pmatrix} \begin{pmatrix} 10 \\ 2 \\ \end{pmatrix}$$

(c) How many committees are possible if a committee must consist of all women or all men? $$\begin{pmatrix} 32 \\ 5 \\ \end{pmatrix} + \begin{pmatrix} 10 \\ 5 \\ \end{pmatrix}$$

(d) How many committees are possible if a committee must have exactly one women? - $$\begin{pmatrix} 32 \\ 1 \\ \end{pmatrix} \begin{pmatrix} 10 \\ 4 \\ \end{pmatrix}$$

(e) How many committees of five different executive positions are possible? (e.g., chair, treasurer, secretary, etc.) - (42)(41)(40)(39)(38)

Thanks for any help!

• Selections for a committee are not ordered. You should be using combinations rather than permutations. On the other hand, the last question is an ordered selection, so you should be using permutations. – N. F. Taussig Oct 31 '15 at 1:48
• a)-d) look fine to me. – callculus Oct 31 '15 at 2:03
• @calculus Selections to a committee are not ordered. Each of those answers should be divided by the number of orders in which the same group of people can be selected. – N. F. Taussig Oct 31 '15 at 2:05
• Suppose you wanted to form a committee of size five but had just five people to choose from, instead of $42$. Would the answer be $(5)(4)(3)(2)(1)$? – Barry Cipra Oct 31 '15 at 2:08
• Ah okay, I see what you're saying. How would this be corrected? Or am I completely off base here – smitty werbenjagermanjensen Oct 31 '15 at 2:11

In part (e), order matters. Selecting Angela to be chair and Barbara to be treasurer is different from selecting Barbara to be chair and Angela to be treasurer. Therefore, there are $42$ ways to fill the position of chair, $41$ ways to pick the treasurer, $40$ ways to select the secretary, and so forth. Hence, there are $$42 \cdot 41 \cdot 40 \cdot 39 \cdot 38 = \frac{42!}{(42 - 5)!} = \frac{42!}{37!}$$ ways to select a committee consisting of five different executive positions. This is a permutation in which $5$ members are selected from a set with $42$ members. The number of ordered selections of $k$ elements from a set of $n$ elements is $$P(n, k) = \frac{n!}{(n - k)!}$$ since there are $n$ ways to make the first selection, $n - 1$ ways to make the second selection, and so forth until there are $n - k + 1$ to make the $k$th selection and $$n(n - 1) \cdots (n - k + 1) = \frac{n(n - 1) \cdots (n - k + 1) \cdot (n - k)!}{(n - k)!} = \frac{n!}{(n - k)!}$$

Unless members of a committee are assigned to specific posts, the order in which the committee is selected does not matter. Selecting Alice, Benjamin, Charles, Diana, and Evelyn to serve on the committee produces the same committee as selecting Diana, Charles, Evelyn, Alice, and Benjamin. There are $5! = 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1$ orders in which we could select the same five people. Since there are $$42 \cdot 41 \cdot 40 \cdot 39 \cdot 38 = \frac{42!}{(42 - 5)!}$$ ways to make an ordered selection, the number of ways we can select a committee of $5$ people from the $42$ members of the club is $$\frac{42 \cdot 41 \cdot 40 \cdot 39 \cdot 38}{5 \cdot 4 \cdot 3 \cdot 2 \cdot 1} = \frac{42!}{(42 - 5)!5!} = \frac{42!}{37!5!}$$ ways to select a committee of five people from the club. This is a combination in which $5$ members are selected from a set with $42$ members. The number of unordered selections (subsets) of $k$ elements from a set with $n$ elements is $$\binom{n}{k} = C(n, k) = \frac{n!}{k!(n - k)!}$$ where we divide the number of ordered selections of $k$ elements by the $k!$ orders in which those elements could be selected.

See if you can revise your answers to parts (b), (c), and (d).

I believe b would be: $C(32, 3)*C(10, 2)$

I encountered this type of problem and tried solving it the way that you are. You are answering b as if order matters, but it doesn't, so use combinations. Similar problem with d, so I'd say $C(32, 1)*C(10, 4)$.

All the last question is asking is: "how many possibilities are there if each position is distinguishable", so $42!/37!$.

42 is 10 men + 32 women