2020 AMC 8 Problems/Problem 23
Contents
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?
Solution 1 (complementary counting)
Without the restriction that each student receives at least one award, we could simply take each of the awards and choose one of the
students to give it to, so that there would be
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
choices for which student that is, then
ways of choosing a student to receive each of the awards, for a total of
. However, if
students both don't receive an award, then such a case would be counted twice among our
, so we need to add back in these cases. Of course,
students both not receiving an award is equivalent to only
student receiving all
awards, so there are simply
choices for which student that would be. It follows that the total number of ways of distributing the awards is
.
Solution 2
Firstly, observe that it is not possible for a single student to receive or
awards because this would mean that one of the other students receives no awards. Thus, each student must receive either
,
, or
awards. If a student receives
awards, then the other two students must each receive
award; if a student receives
awards, then another student must also receive
awards and the remaining student must receive
award. We consider each of these two cases in turn. If a student receives three awards, there are
ways to choose which student this is, and
ways to give that student
out of the
awards. Next, there are
students left and
awards to give out, with each student getting one award. There are clearly just
ways to distribute these two awards out, giving
ways to distribute the awards in this case.
In the other case, a student receives awards. We first have to choose which of the two students we will select to give two awards each to. There are
ways to do this, after which there are
ways to give the first student his two awards, leaving
awards yet to distribute. There are then
ways to give the second student his
awards. Finally, there is only
student and
award left, so there is only
way to distribute this award. This results in
ways to distribute the awards in this case. Adding the results of these two cases, we get
.
Solution 3 (variation of Solution 2)
If each student must receive at least one award, then, as in Solution 2, we deduce that the only possible ways to split up the awards are
and
(i.e. one student gets three awards and the others get one each, or two students each get two awards and the other student is left with the last one). In the first case, there are
choices for which student gets
awards, and
choices for which
awards they get. We are then left with
awards, and there are exactly
choices depending on which remaining student gets which. This yields a total for this case of
. For the second case, there are similarly
choices for which student gets only
award, and
choices for which award he gets. There are then
remaining awards, from which we choose
to give to one student and
to give to the other, which can be done in
ways (and we can say that e.g. the
chosen this way go to the first remaining student and the other
go to the second remaining student, which counts all possibilities). This means the total for the second case is
, and the answer is
.
See also
2020 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 22 |
Followed by Problem 24 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
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