2020 AMC 8 Problems/Problem 23

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?

$\textbf{(A) }120 \qquad \textbf{(B) }150 \qquad \textbf{(C) }180 \qquad \textbf{(D) }210 \qquad \textbf{(E) }240$

Solution 1

Credits to asbodke for the solution: Without the restriction that each student receives one award, the answer is $3^5=243$ . We then subtract the cases where one student doesn't receive an award. There are 3 students to choose from, and then after that, there are $2^5=32$ options, so we subtract $3\cdot32=96$ options. However, the cases where one student receives the awards were counted once originally, but then subtracted twice! However, we only want them to be counted zero times, so we have to add them back again. There are 3 cases when this happens (since there are 3 students). Thus, the answer is $243-96+3=150\implies\boxed{\textbf{(B)}150}$ .

Solution 2

First, observe that it is not possible for a single student to receive four or five awards because this would mean that one of the other students receives no awards. Thus, each student must receive either one, two, or three awards. If a student receives three awards, then the other two students must each receive one award. If a student receives two awards, then another student must also receive two awards, and the remaining student must receive one award. Thus, there are only two cases to consider. Note that the two cases listed below are mutually exclusive and together cover all possible situations.

Case 1: A student receives three awards
First we have to choose which of the three students we will select to give three awards to. There are 3 ways to do this. There are $\binom{5}{3}$ ways to give that student three of the five awards. Next, there are two students left and two awards to give out, with each student getting one award. There are 2 ways to give these two awards out. This results in $3\cdot\binom{5}{3}\cdot 2=60$ ways to distribute the awards in this case.

Case 2: A student receives two awards
First we have to choose which of the two students we will select to give two awards each to. There are $\binom{3}{2}$ ways to do this. There are $\binom{5}{2}$ ways to give the first student his two awards, leaving 3 awards yet to distribute. There are $\binom{3}{2}$ ways to give the second student his two awards. Finally, there is only one student and one award left so there is only one way to distribute this award. This results in $\binom{3}{2}\cdot\binom{5}{2}\cdot\binom{3}{2}\cdot 1=90$ ways to distribute the awards in this case.

Adding the results of these two cases, we get $60+90=150 \implies\boxed{\textbf{(B) }150}$ .
~junaidmansuri

Solution 3

We can distribute the awards in a $3-1-1$ or $1-2-2$ . We will handle each case separately. For the first case, there are $\binom53 \cdot \binom21 \cdot \binom11 \cdot \frac{3!}{2!} = 60$ ways to distribute the prizes.

For the second case, there are $\binom52 \cdot \binom32 \cdot \binom11 \cdot \frac{3!}{2!} = 90$ ways to distribute the prizes.

Therefore, the answer is $60 + 90 = \textbf{(B) }150$ .

-franzliszt

2020 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 22	Followed by Problem 24
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2020 AMC 8 Problems/Problem 23

Contents

Solution 1

Solution 2

Solution 3

See also