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

(Solution 4 (Combinatorial Argument): Deleted my inefficient solution. I will edit this page later.)
(Solution 1 (Without Words): I will push the bash solution to the end.)
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<math>\textbf{(A) } 47\qquad\textbf{(B) } 48\qquad\textbf{(C) } 83\qquad\textbf{(D) } 95\qquad\textbf{(E) } 96\qquad</math>
 
<math>\textbf{(A) } 47\qquad\textbf{(B) } 48\qquad\textbf{(C) } 83\qquad\textbf{(D) } 95\qquad\textbf{(E) } 96\qquad</math>
 
==Solution 1 (Without Words)==
 
<cmath>\begin{array}{c|c|c}
 
\text{Tenors} & \text{Basses} & \text{Groups} \
 
\hline
 
0 & 4, 8 & \binom{6}{0}\left[\binom{8}{4}+\binom{8}{8}\right]=70+1=\fbox{71} \
 
1 & 1, 5 & \binom{6}{1}\left[\binom{8}{1}+\binom{8}{5}\right]=6(8+56)=6(64)=3\fbox{84} \
 
2 & 2, 6 & \binom{6}{2}\left[\binom{8}{2}+\binom{8}{6}\right]=15(28+28)=15(56)=8\fbox{40} \
 
3 & 3, 7 & \binom{6}{3}\left[\binom{8}{3}+\binom{8}{7}\right]=20(56+8)=20(64)=12\fbox{80} \
 
4 & 0, 4, 8 & \binom{6}{4}\left[\binom{8}{0}+\binom{8}{4}+\binom{8}{8}\right]=15(1+70+1)=15(72)=10\fbox{80} \
 
5 & 1, 5 & \binom{6}{5}\left[\binom{8}{1}+\binom{8}{5}\right]=6(8+56)=6(64)=3\fbox{84} \
 
6 & 2, 6 & \binom{6}{6}\left[\binom{8}{2}+\binom{8}{6}\right]=1(28+28)=\fbox{56}
 
\end{array}</cmath>
 
 
<math>71+384+840+1280+1080+384+56=40\boxed{\textbf{(D)} ~95}</math>
 
  
 
==Solution 2 (Generating Functions)==
 
==Solution 2 (Generating Functions)==

Revision as of 22:49, 23 August 2021

Problem

A choir director 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 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)

By casework, we construct the following table. In the last column, we rewrite some of the combinations using the identity $\binom{n}{r}=\binom{n}{n-r}:$ \[\begin{array}{c|c|c|c} & & & \\ [-2ex] \textbf{\# of Tenors} & \textbf{\# of Basses} & \textbf{\# of Ways} & \textbf{Rewrite \# of Ways} \\ [0.5ex] \hline\hline  & & & \\ [-2ex] 0 & 8 & \tbinom{6}{0}\tbinom{8}{8} & \\ [1ex] 1 & 1 & \tbinom{6}{1}\tbinom{8}{1} & \tbinom{6}{1}\tbinom{8}{7}\\ [1ex] 2 & 2 & \tbinom{6}{2}\tbinom{8}{2} & \tbinom{6}{2}\tbinom{8}{6}\\ [1ex] 3 & 3 & \tbinom{6}{3}\tbinom{8}{3} & \tbinom{6}{3}\tbinom{8}{5}\\ [1ex] 4 & 4 & \tbinom{6}{4}\tbinom{8}{4} & \\ [1ex] 5 & 5 & \tbinom{6}{5}\tbinom{8}{5} & \tbinom{6}{5}\tbinom{8}{3}\\ [1ex] 6 & 6 & \tbinom{6}{6}\tbinom{8}{6} & \tbinom{6}{6}\tbinom{8}{2}\\ [1ex] \hline & & & \\ [-2ex] 0 & 4 & \tbinom{6}{0}\tbinom{8}{4} & \\ [1ex] 1 & 5 & \tbinom{6}{1}\tbinom{8}{5} & \tbinom{6}{1}\tbinom{8}{3}\\ [1ex] 2 & 6 & \tbinom{6}{2}\tbinom{8}{6} & \tbinom{6}{2}\tbinom{8}{2}\\ [1ex] 3 & 7 & \tbinom{6}{3}\tbinom{8}{7} & \tbinom{6}{3}\tbinom{8}{1}\\ [1ex] 4 & 8 & \tbinom{6}{4}\tbinom{8}{8} & \tbinom{6}{4}\tbinom{8}{0}\\ [1ex] \hline & & & \\ [-2ex] 4 & 0 & \tbinom{6}{4}\tbinom{8}{0} & \tbinom{6}{2}\tbinom{8}{0}\\ [1ex] 5 & 1 & \tbinom{6}{5}\tbinom{8}{1} & \tbinom{6}{1}\tbinom{8}{1}\\ [1ex] 6 & 2 & \tbinom{6}{6}\tbinom{8}{2} & \tbinom{6}{0}\tbinom{8}{2}\\ [1ex] \end{array}\] We apply Vandermonde's Identity to find the requested sum: \begin{align*} N&=\underbrace{\left[\sum_{k=0}^{6}\binom{6}{k}\binom{8}{8-k}\right]}_{\tbinom{14}{8}}+\underbrace{\left[\sum_{k=0}^{4}\binom{6}{k}\binom{8}{4-k}\right]}_{\tbinom{14}{4}}+\underbrace{\left[\sum_{k=0}^{2}\binom{6}{k}\binom{8}{2-k}\right]}_{\tbinom{14}{2}} \\ &=\binom{14}{6}+\binom{14}{4}+\binom{14}{2} \\ &=3003+1001+91 \\ &=4095 \\ &\equiv\boxed{\textbf{(D) } 95}\pmod{100}. \end{align*}

~MRENTHUSIASM

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

See also

2021 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 14
Followed by
Problem 16
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All AMC 12 Problems and Solutions

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