2010 AMC 10B 第 22 题

Question

七块不同的糖果要分到三个袋子中。红袋和蓝袋必须各至少得到一块糖，白袋可以为空。有多少种分法？

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

答案：C

知识点：补集计数容斥原理乘法原理

难度评级：1790

小提示：

先数所有

3^7

种分法。

Count all

3^7

distributions first

大提示：

减去红袋或蓝袋为空的情况，再加回重叠。

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

解答：

可以用补集计数。无限制地分配糖果的方法总数为

3^7 = 2187\text{。}

要数无效安排，需要考虑红袋或蓝袋为空的情况。

若红袋为空，则每块糖只有

2

个袋子可选，共

2^7 = 128

种安排。蓝袋为空也同样有

128

种。

这两类有一个重叠情况：红袋和蓝袋都为空。因此有效分法的数量为

2187 - (128 + 128 - 1) = 1932\text{。}

所以正确答案是 C。

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.

2010 AMC 10B 第 22 题

答案：C

难度评级：1790

小提示：

大提示：

解答：

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