Difference between revisions of "2020 AMC 8 Problems/Problem 23"

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Problem

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 (Principle of Inclusion-Exclusion)

Without the restriction that each student receives at least one award, we could simply take each of the $5$ awards and choose one of the $3$ students to give it to, so that there would be $3^5=243$ ways to distribute the awards. We now need to subtract the cases where at least one student doesn't receive an award. If a student doesn't receive an award, there are $3$ choices for which student that is, then $2^5 = 32$ ways of choosing a student to receive each of the awards, for a total of $3 \cdot 32 = 96$ . However, if $2$ students both don't receive an award, then such a case would be counted twice among our $96$ , so we need to add back in these cases. Of course, $2$ students both not receiving an award is equivalent to only $1$ student receiving all $5$ awards, so there are simply $3$ choices for which student that would be. Therefore, the total number of ways of distributing the awards is $243-96+3=\boxed{\textbf{(B) }150}$ .

Solution 2 (Constructive Counting)

Firstly, observe that it is not possible for a single student to 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 student receives $2$ awards and one student recieves $1$ 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 distribute. 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 $60+90=\boxed{\textbf{(B) }150}$ .

Video Solution by WhyMath

https://youtu.be/HkFQe7ZxBb4

~WhyMath

Video Solutions by The Learning Royal

https://youtu.be/tDChKU0pVN4

https://youtu.be/RUg6QfV5yg4

Video Solution by Interstigation

https://youtu.be/YnwkBZTv5Fw?t=1443

~Interstigation

Video Solution by STEMbreezy

https://youtu.be/wq8EUCe5oQU?t=243

~STEMbreezy

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