2010 AMC 10B Problems/Problem 22

Problem

Seven distinct pieces of candy are to be distributed among three bags. The red bag and the blue bag must each receive at least one piece of candy; the white bag may remain empty. How many arrangements are possible?

$\mathrm{(A)}\ 1930 \qquad \mathrm{(B)}\ 1931 \qquad \mathrm{(C)}\ 1932 \qquad \mathrm{(D)}\ 1933 \qquad \mathrm{(E)}\ 1934$

Solution

We can count the total number of ways to distribute the candies (ignoring the restrictions), and then subtract the overcount to get the answer.

Each candy has three choices; it can go in any of the three bags.

Since there are seven candies, that makes the total distributions $3^7=2187$

To find the overcount, we calculate the number of invalid distributions: the red or blue bag is empty.

The number of distributions such that the red bag is empty is equal to $2^7$ , since it's equivalent to distributing the 7 candies into 2 bags.

We know that the number of distributions with the blue bag is empty will be the same number because of the symmetry, so it's also $2^7$ .

The case where both the red and the blue bags are empty (all 7 candies are in the white bag) are included in both of the above calculations, and this case has only $1$ distribution.

The total overcount is $2^7+2^7-1=2^8-1$

The final answer will be $\text{total}-\text{overcount}=2187-(2^8-1) = 2187-256+1=1931+1=\boxed{1932}$

That makes the letter choice C

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2010 AMC 10B Problems/Problem 22

Problem

Solution