2020 AMC 8 第 23 题

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所有题目均经美国数学协会(MAA)官方合法授权使用。

23.

五个不同的奖项要颁给三名学生。每名学生至少获得一个奖项。共有多少种不同的颁奖方式?

Five different awards are to be given to three students. Each student will receive at least one award. In how many different ways can the awards be distributed?

120120

150150

180180

210210

240240

答案:B
知识点:容斥原理乘法原理
难度评级:1370
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文字解答:

若没有“每人至少一个奖项”的限制,每个奖项有 33 个学生可选,55 个奖项共有 35=2433^5=243 种分配。

减去至少一名学生没有获奖的情况。指定某名学生没有获奖时,五个奖项只能给另外两人,有 252^5 种;三名学生都可能是没有获奖者,所以先减去 3253\cdot2^5。不过所有奖项都给同一个人的 33 种情况被多减了一次。

由容斥,符合条件的分配数为 35325+3=1503^5-3\cdot2^5+3=150

正确答案是 B

There are 35=2433^5=243 ways to give each of the 55 distinct awards to one of the 33 students.

Subtract the distributions in which at least one student receives no award. If a particular student receives none, the awards go to the other two students in 252^5 ways. This gives 3253\cdot2^5 counts, but the 33 cases in which one student receives all awards have each been subtracted twice.

By inclusion-exclusion, the desired number is 35325+3=150.3^5-3\cdot2^5+3=150.

Thus, the correct answer is B.

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