2010 AMC 10B Problem 22

Attempt Problem 22 of the 2010 AMC 10B below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2010 AMC 10B solutions, or check the answer key.

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22.

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?

19301930

19311931

19321932

19331933

19341934

Answer: C
Concepts:complementary countinginclusion-exclusionmultiplication principle
Difficulty rating: 1790
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.

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