Difference between revisions of "2021 AMC 12A Problems/Problem 15"

(Video Solution (using Vandermonde's Identity))
(Video Solution by pi_is_3.14 (using Vandermonde's Identity))
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https://www.youtube.com/watch?v=AjQARBvdZ20
 
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==Video Solution by pi_is_3.14 (using Vandermonde's Identity)==
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==Video Solution by OmegaLearn (using Vandermonde's Identity)==
 
https://www.youtube.com/watch?v=mki7xtZLk1I
 
https://www.youtube.com/watch?v=mki7xtZLk1I
  

Revision as of 23:13, 11 February 2021

Problem

A choir direction must select a group of singers from among his $6$ tenors and $8$ basses. The only requirements are that the difference between the number of tenors and basses must be a multiple of $4$, and the group must have at least one singer. Let $N$ be the number of different groups that could be selected. What is the remainder when $N$ is divided by $100$?

$\textbf{(A) } 47\qquad\textbf{(B) } 48\qquad\textbf{(C) } 83\qquad\textbf{(D) } 95\qquad\textbf{(E) } 96\qquad$

Video Solution by Punxsutawney Phil

https://youtube.com/watch?v=FD9BE7hpRvg&t=533s

Solution 2

We know the choose function and we know the pair multiplication $MN$ so we do the multiplications and additions. $\binom{6}{0}(\binom{8}{4}+\binom{8}{8})+\binom{6}{1}(\binom{8}{1}+\binom{8}{5})+\binom{6}{2}(\binom{8}{2}+\binom{8}{6})+\binom{6}{3}(\binom{8}{3}+\binom{8}{7})+\binom{6}{4}(\binom{8}{0}+\binom{8}{4}+\binom{8}{8})+\binom{6}{5}(\binom{8}{1}+\binom{8}{5})+\binom{6}{6}(\binom{8}{2}+\binom{8}{6}) = \boxed{(D) 4095}$ ~Lopkiloinm

Video Solution by Hawk Math

https://www.youtube.com/watch?v=AjQARBvdZ20

Video Solution by OmegaLearn (using Vandermonde's Identity)

https://www.youtube.com/watch?v=mki7xtZLk1I

~pi_is_3.14

See also

2021 AMC 12A (ProblemsAnswer KeyResources)
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

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