Difference between revisions of "2020 AMC 8 Problems/Problem 23"
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Without the restriction that each student receives at least one award, we could simply take each of the <math>5</math> awards and choose one of the <math>3</math> students to give it to, so that there would be <math>3^5=243</math> 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 <math>3</math> choices for which student that is, then <math>2^5 = 32</math> ways of choosing a student to receive each of the awards, for a total of <math>3 \cdot 32 = 96</math>. However, if <math>2</math> students both don't receive an award, then such a case would be counted twice among our <math>96</math>, so we need to add back in these cases. Of course, <math>2</math> students both not receiving an award is equivalent to only <math>1</math> student receiving all <math>5</math> awards, so there are simply <math>3</math> choices for which student that would be. It follows that the total number of ways of distributing the awards is <math>243-96+3=\boxed{\textbf{(B) }150}</math>. | Without the restriction that each student receives at least one award, we could simply take each of the <math>5</math> awards and choose one of the <math>3</math> students to give it to, so that there would be <math>3^5=243</math> 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 <math>3</math> choices for which student that is, then <math>2^5 = 32</math> ways of choosing a student to receive each of the awards, for a total of <math>3 \cdot 32 = 96</math>. However, if <math>2</math> students both don't receive an award, then such a case would be counted twice among our <math>96</math>, so we need to add back in these cases. Of course, <math>2</math> students both not receiving an award is equivalent to only <math>1</math> student receiving all <math>5</math> awards, so there are simply <math>3</math> choices for which student that would be. It follows that the total number of ways of distributing the awards is <math>243-96+3=\boxed{\textbf{(B) }150}</math>. | ||
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==Solution 3 (variation of Solution 2)== | ==Solution 3 (variation of Solution 2)== |
Revision as of 16:40, 1 May 2021
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 .
This is confusing, so you do not need to know...
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 .
Video Solution by WhyMath
~savannahsolver
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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