2020 AMC 8 Problems/Problem 23
Contents
- 1 Problem
- 2 Solution 1 (Principle of Inclusion-Exclusion)
- 3 Solution 2 (Constructive Counting)
- 4 Solution 3 (Casework)
- 5 Video Solution
- 6 Video Solution by OmegaLearn
- 7 Video Solution by SpreadTheMathLove
- 8 Video Solution by WhyMath
- 9 Video Solutions by The Learning Royal
- 10 Video Solution by Interstigation
- 11 Video Solution by STEMbreezy
- 12 See also
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 (Principle of Inclusion-Exclusion)
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. Therefore, the total number of ways of distributing the awards is .
Solution 2 (Constructive 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 choices for which student gets 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 .
Solution 3 (Casework)
Upon inspection (specified in the above solution), there are two cases of the distribution of awards to the students: one student gets 3 awards and the other each get 1 award or one student gets 1 award and the other two get 2 awards.
In the first case, there are ways to choose the person who gets 3 awards. From here, there are ways to choose the 3 awards from the 5 total awards. Now, one person has choices for awards and the other has choice for the award. Thus, the total number of ways to choose awards in this case is .
In the other case, there are ways to choose the person who gets 1 award, and choices for his/her award. Then, one person has ways to have his/her awards and the other person has ways to have his/her awards. This gives ways for this case.
Adding these cases together, we get ways to distribute the awards, or choice .
~TaeKim
Video Solution
Please like and subscribe!
Video Solution by OmegaLearn
https://youtu.be/dFFjlxm43b0?t=899
~ pi_is_3.14
Video Solution by SpreadTheMathLove
https://www.youtube.com/watch?v=Dg_2wXNY3tE
Video Solution by WhyMath
~WhyMath
Video Solutions by The Learning Royal
Video Solution by Interstigation
https://youtu.be/YnwkBZTv5Fw?t=1443
~Interstigation
Video Solution by STEMbreezy
https://youtu.be/wq8EUCe5oQU?t=243
~STEMbreezy
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.