2020 AMC 8

Complete problem set with solutions and individual problem pages

Problem 23 Hard

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?

  • A.

    120

  • B.

    150

  • C.

    180

  • D.

    210

  • E.

    240

Answer:B

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}.