2008 AMC 10A Problem 23

Call the two subsets $X$ and $Y$.

1. Where can each letter go? Together $X$ and $Y$ must contain all of $a, b, c, d, e$, so no letter is left out. Each letter is in exactly one of three places: in both, in $X$ only, or in $Y$ only.
2. The shared letters. Exactly two letters are in both subsets. The number of ways to choose $2$ letters from $5$ is
$$\frac{5 \times 4}{2} = 10.$$
3. The other three letters. Each of them goes to $X$ only or to $Y$ only, so there are $2 \times 2 \times 2 = 8$ choices.
4. So far that is $10 \times 8 = 80$ ways, but this treats "first subset" and "second subset" as different.
5. Order does not matter. For example, $X = \{a, b, c\}$, $Y = \{a, b, d, e\}$ and $X = \{a, b, d, e\}$, $Y = \{a, b, c\}$ are the same choice, and we counted it twice. Every choice is counted exactly twice, because $X$ and $Y$ are never equal (the three unshared letters are in only one of them). So divide by $2$:
$$\frac{80}{2} = 40.$$

The answer is $\boxed{\textbf{(B)}\ 40}$.

Choosing two subsets is the same as deciding, for each letter, which part of a two-circle Venn diagram it goes in. That turns a set question into a "multiply the choices" question. Whenever order does not matter, count with order first and then divide, but first check that no choice is counted only once.

Avoid the division. After picking the $2$ shared letters ($10$ ways), you only need to split the other $3$ letters into two groups, one for each subset. There are $8$ ways to pick which letters join $X$, but picking $\{c\}$ for $X$ and picking $\{d, e\}$ for $X$ give the same pair of subsets. So there are only $8 \div 2 = 4$ different splits, for a total of $10 \times 4 = 40$.