AMC8 2020

AMC8 2020 · Q23

AMC8 2020 · Q23. It mainly tests Combinations, Casework.

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?

要将五个不同的奖项颁发给三个学生。每个学生至少获得一个奖项。奖项可以以多少种不同的方式分配？

(A) 120 120

(B) 150 150

(D) 210 210

(E) 240 240

Answer

Correct choice: (B)

正确答案：（B）

Solution

Firstly, observe that a single student can't receive $4$ or $5$ awards because this would mean that one of the other students receives no awards. Thus, each student must receive either $1$, $2$, or $3$ awards. If a student receives $3$ awards, then the other two students must each receive $1$ award; if a student receives $2$ awards, then another student must also receive $2$ awards and the remaining student must receive $1$ award. We consider each of these two cases in turn. If a student receives three awards, there are $3$ ways to choose which student this is, and $\binom{5}{3}$ ways to give that student $3$ out of the $5$ awards. Next, there are $2$ students left and $2$ awards to give out, with each student getting one award. There are clearly just $2$ ways to distribute these two awards out, giving $3\cdot\binom{5}{3}\cdot 2=60$ ways to distribute the awards in this case. In the other case, two students receive $2$ awards and one student receives $1$ award. We know there are $3$ choices for which student gets $1$ award. There are $\binom{3}{1}$ ways to do this. Then, there are $\binom{5}{2}$ ways to give the first student his two awards, leaving $3$ awards yet to be distributed. There are then $\binom{3}{2}$ ways to give the second student his $2$ awards. Finally, there is only $1$ student and $1$ award left, so there is only $1$ way to distribute this award. This results in $\binom{5}{2}\cdot\binom{3}{2}\cdot 1\cdot 3 =90$ ways to distribute the awards in this case. Adding the results of these two cases, we get $60+90=\boxed{\textbf{(B) }150}$.

首先，观察到一个学生不能获得 $4$ 或 $5$ 个奖项，因为这意味着其他学生中有一个得不到奖项。因此，每个学生必须获得 $1$、$2$ 或 $3$ 个奖项。如果一个学生获得 $3$ 个奖项，则另外两个学生各获得 $1$ 个；如果一个学生获得 $2$ 个，则另一个学生也必须获得 $2$ 个，剩余学生获得 $1$ 个。我们分别考虑这两种情况。如果一个学生获得三个奖项，有 $3$ 种选择决定是哪个学生，然后 $\binom{5}{3}$ 种方式给他 $5$ 个奖项中的 $3$ 个。接下来，剩下 $2$ 个学生和 $2$ 个奖项，每个学生得一个，显然只有 $2$ 种分配方式，因此本种情况有 $3\cdot\binom{5}{3}\cdot 2=60$ 种方式。另一种情况，两个学生各得 $2$ 个奖项，一个学生得 $1$ 个。有 $3$ 种选择决定哪个学生得 $1$ 个奖项。然后，有 $\binom{5}{2}$ 种方式给第一个学生两个奖项，剩下 $3$ 个奖项。然后有 $\binom{3}{2}$ 种方式给第二个学生两个奖项。最后，只剩 $1$ 个学生和 $1$ 个奖项，只有 $1$ 种方式。因此本种情况有 $\binom{5}{2}\cdot\binom{3}{2}\cdot 1\cdot 3 =90$ 种方式。两种情况总计 $60+90=\boxed{\textbf{(B) }150}$。

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