2010 AMC 10B Problem 22

The candies are all different, so an arrangement is just a choice of bag for each candy. It is easier to count all arrangements and take away the bad ones.

Warm-up with $2$ candies. Each candy has $3$ choices, so $3 \cdot 3 = 9$ arrangements. Red empty: each candy goes to blue or white, $2 \cdot 2 = 4$ ways. Blue empty: also $4$. But "both candies in white" is in both groups of $4$. So the bad arrangements number $4 + 4 - 1 = 7$, and the good ones $9 - 7 = 2$: candy one in red and candy two in blue, or the other way round. That is correct.

Now $7$ candies.

1. All arrangements: each of the $7$ candies has $3$ choices, $3^7 = 2187$.
2. Red bag empty: each candy goes to blue or white, $2^7 = 128$ ways.
3. Blue bag empty: likewise $128$ ways.
4. Both red and blue empty: every candy in white, $1$ way. This arrangement was taken away in step 2 and again in step 3, so add it back once.
5. Good arrangements:
$$2187 - 128 - 128 + 1 = 1932.$$

The answer is $\boxed{\textbf{(C)}\ 1932}$.

For "each bag gets at least one" rules, count everything and subtract the arrangements that break a rule. When two rules can be broken at the same time, those arrangements get subtracted twice, so add them back once. The answer choices differ by $1$ exactly to catch a missing add-back.

Count the good arrangements directly. Decide how many candies, $k$, go into red and blue together ($k$ from $2$ to $7$); the rest go into white. Choose those $k$ candies in $\binom{7}{k}$ ways. Each of them goes to red or blue, $2^k$ ways, minus the $2$ ways that put them all in one bag.

The total is $42 + 210 + 490 + 630 + 434 + 126 = 1932$.