Difference between revisions of "2020 AMC 8 Problems/Problem 23"
Icematrix2 (talk | contribs) |
(→Solution 1) |
||
| Line 5: | Line 5: | ||
==Solution 1== | ==Solution 1== | ||
| − | <math>\boxed{\textbf{(B) }150}</math> | + | Credits to asbodke for the solution: |
| + | Without the restriction that each student receives one award, the answer is <math>3^5=243</math>. We then subtract the cases where one student doesn't receive an award. There are 3 students to choose from, and then after that, there are <math>2^5=32</math> options, so we subtract <math>3\cdot32=96</math> options. However, the cases where one student receives the awards were counted once originally, but then subtracted twice! However, we only want them to be counted zero times, so we have to add them back again. There are 3 cases when this happens (since there are 3 students). Thus, the answer is <math>243-96+3=150\implies\boxed{\textbf{(B)}150}</math>. | ||
| + | |||
==See also== | ==See also== | ||
{{AMC8 box|year=2020|num-b=22|num-a=24}} | {{AMC8 box|year=2020|num-b=22|num-a=24}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 02:32, 18 November 2020
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
Credits to asbodke for the solution:
Without the restriction that each student receives one award, the answer is 
. We then subtract the cases where one student doesn't receive an award. There are 3 students to choose from, and then after that, there are 
options, so we subtract 
options. However, the cases where one student receives the awards were counted once originally, but then subtracted twice! However, we only want them to be counted zero times, so we have to add them back again. There are 3 cases when this happens (since there are 3 students). Thus, 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 | ||
| All AJHSME/AMC 8 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
