2021 AMC 12A Problems/Problem 15

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

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Page

Toolbox

Search

2021 AMC 12A Problems/Problem 15

Contents

Problem

Video Solution by Punxsutawney Phil

Solution 2

Video Solution by Hawk Math

See also