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

Revision as of 17:43, 13 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 $100$ ?

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

Solution 1

We know the choose function and we know the pair multiplication $MN$ so we do the multiplications and additions. $\binom{6}{0}\left(\binom{8}{4}+\binom{8}{8}\right)+\binom{6}{1}\left(\binom{8}{1}+\binom{8}{5}\right)+\binom{6}{2}\left(\binom{8}{2}+\binom{8}{6}\right)+\binom{6}{3}\left(\binom{8}{3}+\binom{8}{7}\right)+\\\binom{6}{4}\left(\binom{8}{0}+\binom{8}{4}+\binom{8}{8}\right)+\binom{6}{5}\left(\binom{8}{1}+\binom{8}{5}\right)+\binom{6}{6}\left(\binom{8}{2}+\binom{8}{6}\right) = 4095=\equiv\boxed{(D) 95}\pmod{100}$

~Lopkiloinm

Solution 2 (Generating Functions)

The problem can be done using a roots of unity filter. Let $f(x,y)=(1+x)^8(1+y)^6$ . By expanding the binomials and distributing, $f(x,y)$ is the generating function for different groups of basses and tenors. That is, $f(x,y)=\sum_{m=0}^8\sum_{n=0}^6 a_{mn}x^my^n$ where $a_{mn}$ is the number of groups of $m$ basses and $n$ tenors. What we want to do is sum up all values of $a_{mn}$ for which $4\mid m-n$ except for $a_{00}=1$ . To do this, define a new function $g(x)=f(x,x^{-1})=\sum_{m=0}^8\sum_{n=0}^6 a_{mn}x^{m-n}=(1+x)^8(1+x^{-1})^6.$ Now we just need to sum all coefficients of $g(x)$ for which $4\mid m-n$ . Consider a monomial $h(x)=x^k$ . If $4\mid k$ , $h(i)+h(-1)+h(-i)+h(1)=1+1+1+1=4$ otherwise, $h(i)+h(-1)+h(-i)+h(1)=0.$ $g(x)$ is a sum of these monomials so this gives us a method to determine the sum we're looking for: $\frac{g(i)+g(-1)+g(-i)+g(1)}{4}=2^{12}=4096$ (since $g(-1)=0$ and it can be checked that $g(i)=-g(-i)$ ). Hence, the answer is $4096-1$ with the $-1$ for $a_{00}$ which gives $\boxed{95}$ . ~lawliet163

Solution 3 (Casework and Vandermonde's Identity)

$\begin{array}{cccc} \text{\# of Tenors} & \text{\# of Basses} & \text{\# of Ways} & \text{Rewrite \# of Ways} \\ \hline\hline 1 & 1 & \binom{6}{1}\binom{8}{1} & \\ 2 & 2 & \binom{6}{2}\binom{8}{2} & \\ 3 & 3 & \binom{6}{3}\binom{8}{3} & \\ 4 & 4 & 3 & \\ 5 & 5 & 4 &\\ 6 & 6 & 5 &\\ 0 & 8 & 6 &\\ \hline 0 & 4 & 7 &\\ 1 & 5 & 8 &\\ 2 & 6 & 9 &\\ 3 & 7 & 1 &\\ 4 & 8 & 2 &\\ \hline 4 & 0 & 3 &\\ 5 & 1 & 4 &\\ 6 & 2 & 4 & \end{array}$

Video Solution by Punxsutawney Phil

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

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

@@ Line 6: / Line 6: @@
 ==Solution 1==
 We know the choose function and we know the pair multiplication <math>MN</math> so we do the multiplications and additions.
-<math>\binom{6}{0}\left(\binom{8}{4}+\binom{8}{8}\right)+\binom{6}{1}\left(\binom{8}{1}+\binom{8}{5}\right)+\binom{6}{2}\left(\binom{8}{2}+\binom{8}{6}\right)+\binom{6}{3}\left(\binom{8}{3}+\binom{8}{7}\right)+\\\binom{6}{4}\left(\binom{8}{0}+\binom{8}{4}+\binom{8}{8}\right)+\binom{6}{5}\left(\binom{8}{1}+\binom{8}{5}\right)+\binom{6}{6}\left(\binom{8}{2}+\binom{8}{6}\right) = \boxed{(D) 4095}</math>
+<math>\binom{6}{0}\left(\binom{8}{4}+\binom{8}{8}\right)+\binom{6}{1}\left(\binom{8}{1}+\binom{8}{5}\right)+\binom{6}{2}\left(\binom{8}{2}+\binom{8}{6}\right)+\binom{6}{3}\left(\binom{8}{3}+\binom{8}{7}\right)+\\\binom{6}{4}\left(\binom{8}{0}+\binom{8}{4}+\binom{8}{8}\right)+\binom{6}{5}\left(\binom{8}{1}+\binom{8}{5}\right)+\binom{6}{6}\left(\binom{8}{2}+\binom{8}{6}\right) = 4095=\equiv\boxed{(D) 95}\pmod{100}</math>
 ~Lopkiloinm

2021 AMC 12A (Problems • Answer Key • Resources)
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