Difference between revisions of "2021 AMC 12A Problems/Problem 15"
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==Video Solution by Punxsutawney Phil== | ==Video Solution by Punxsutawney Phil== | ||
https://youtube.com/watch?v=FD9BE7hpRvg&t=533s | https://youtube.com/watch?v=FD9BE7hpRvg&t=533s | ||
| + | |||
| + | ==Solution 2== | ||
| + | <math>\binom{6}{0}(\binom{8}{4}+\binom{8}{8})+\binom{6}{1}(\binom{8}{1}+\binom{8}{5})+\binom{6}{2}(\binom{8}{2}+\binom{8}{6})+\binom{6}{3}(\binom{8}{3}+\binom{8}{7})+\binom{6}{4}(\binom{8}{0}+\binom{8}{4}+\binom{8}{8})+\binom{6}{5}(\binom{8}{1}+\binom{8}{5})+\binom{6}{6}(\binom{8}{2}+\binom{8}{6}) = 4095</math> | ||
==See also== | ==See also== | ||
Revision as of 14:30, 11 February 2021
Problem
A choir direction must select a group of singers from among his 
tenors and 
basses. The only requirements are that the difference between the number of tenors and basses must be a multiple of 
, and the group must have at least one singer. Let 
be the number of different groups that could be selected. What is the remainder when is divided by 
?
Video Solution by Punxsutawney Phil
https://youtube.com/watch?v=FD9BE7hpRvg&t=533s
Solution 2
See also
| 2021 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 14 |
Followed by Problem 16 |
| 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 AMC 12 Problems and Solutions | |
| 2021 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 16 |
Followed by Problem 18 | |
| 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 AMC 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
