2010 AMC 10B Problem 22

The candies are distinct, so an arrangement is a choice of bag for each candy: $3^7 = 2187$ assignments with no restriction.

Remove the bad ones by inclusion-exclusion:
- Red empty: each candy goes to blue or white, $2^7 = 128$ ways.
- Blue empty: likewise $128$ ways.
- Both empty: everything in white, $1$ way, which was subtracted twice.

Valid arrangements:
$$2187 - 128 - 128 + 1 = 1932.$$

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

"At least one" restrictions on distinguishable objects are handled by counting the complement: unrestricted assignments minus the ones that violate a condition, corrected for double subtraction. The answer choices differ by one precisely to punish forgetting the add-back term.

Direct count: choose the nonempty red-and-blue portion of $k$ candies ($k$ from $2$ to $7$), split it into two nonempty labeled bags in $2^k - 2$ ways, and put the rest in white: $\sum_{k=2}^{7} \binom{7}{k}(2^k - 2) = 1932$.

• Forgetting to add back the single arrangement with both red and blue empty (all candy in white), which gives 1931.
• Treating the candies as identical and using stars and bars, or subtracting $2^7$ only once.