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2010 AMC 10B Problem 22

Problem 22 of 25HarderCounting & Probability

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?

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Solution

We can count this with complementary counting. The total number of ways to distribute the candies with no restrictions is 37=2187. 3^7 = 2187. To find the number of invalid arrangements, we have to count the number of ways where either the red or blue bag is empty. For the case where the red bag is empty, each candy has 22 options for the bag that goes into. There are then 27=128 2^7 = 128 arrangements for this case. Similarly, there are 128128 arrangements for the case where the blue bag is empty. There is an overlap of one case where both bags are empty. The final answer is then 2187(128+1281)=1932. 2187 - (128 + 128 - 1) = 1932. Thus, C is the correct answer.

More practice

Concepts: complementary counting · inclusion-exclusion · multiplication principle

Problem text and solution from the LIVE past-contest archive. See also the AoPS wiki page for community solutions.