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!
 A: 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).  
A: 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
