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 (someone please rephrase this to make more sense)
Let the rows be , from left to right, and the columns be
, from top to bottom. Notice that the line of three
s can be in either
or
to allow for another line of three
s. In each of those cases, there are two choices for where that line could go (since it has to be oriented the same and in a different row/column), and
choices (triangle or circle for three remaining grid spaces) for the other row/column. However, notice that we overcount the cases in which the chosen row/column in which the line of triangles or circles is identical to the other column in the case that it was, too, all triangles or circles. For this, we subtract
(for rows and columns) cases. Therefore, our answer is
.
Solution 2 (Counting)
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, two student receives awards and one student recieves
award . We know there are 3 choices for which student gets 1 award. There are
ways to do this. Then, 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
.
Video Solution by WhyMath
~WhyMath
Video Solutions
https://youtu.be/tDChKU0pVN4
https://youtu.be/RUg6QfV5yg4
Video Solution by Interstigation
https://youtu.be/YnwkBZTv5Fw?t=1443
~Interstigation
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 | ||
All AJHSME/AMC 8 Problems and Solutions |
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