2010 AMC 10B Problem 22

Question

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?

1930

1931

1932

1933

1934

Answer: C

Concepts:complementary counting inclusion-exclusion multiplication principle

Difficulty rating: 1790

Small hint:

Count all

3^7

distributions first

Big hint:

Subtract cases with the red or blue bag empty, then add back the overlap

Solution:

We can count this with complementary counting. The total number of ways to distribute the candies with no restrictions is

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

2

options for the bag that goes into. There are then

2^7 = 128

arrangements for this case. Similarly, there are

128

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 + 128 - 1) = 1932.

Thus, C is the correct answer.