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

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==Solution 1 (Bijection)==
 
==Solution 1 (Bijection)==
Let <math>t</math> be the number of tenors selected and <math>b</math> be the number of basses selected. The requirements are <math>t\equiv b\pmod{4}</math> and <math>(t,b)\neq(0,0).</math>
+
Suppose that <math>t</math> tenors and <math>b</math> basses will sing. The requirements are <math>t\equiv b\pmod{4}</math> and <math>(t,b)\neq(0,0).</math>
  
It follows that <math>b'=8-b</math> is the number of basses not selected. Since the ordered pairs <math>(t,b)</math> and the ordered pairs <math>(t,b')</math> have one-to-one correspondence, we consider the ordered pairs <math>(t,b')</math> instead. The requirements become <math>t\equiv 8-b'\pmod{4}</math> and <math>(t,8-b')\neq(0,0),</math> which simplify to <math>t+b'\equiv8\pmod{4}</math> and <math>(t,b')\neq(0,8),</math> respectively.
+
It follows that <math>b'=8-b</math> basses will not sing. Since the ordered pairs <math>(t,b)</math> and the ordered pairs <math>(t,b')</math> have one-to-one correspondence, we consider the ordered pairs <math>(t,b')</math> instead. The requirements become <math>t\equiv8-b'\pmod{4}</math> and <math>(t,8-b')\neq(0,0),</math> which simplify to <math>t+b'\equiv0\pmod{4}</math> and <math>(t,b')\neq(0,8),</math> respectively.
 +
 
 +
As <math>t+b'\in\{0,4,6\},</math> the total number of such groups is
 +
<cmath>\begin{align*}
 +
N&=\binom{14}{0}+\binom{14}{4}+\left[\binom{14}{8}-1\right]+\binom{14}{12} \
 +
&=\binom{14}{0}+\binom{14}{4}+\left[\binom{14}{6}-1\right]+\binom{14}{2} \
 +
\end{align*}</cmath>
  
 
==Solution 2 (Generating Functions)==
 
==Solution 2 (Generating Functions)==
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&=\binom{14}{6}+\binom{14}{4}+\binom{14}{2} \
 
&=\binom{14}{6}+\binom{14}{4}+\binom{14}{2} \
 
&=3003+1001+91 \
 
&=3003+1001+91 \
&=4095 \
+
&=4095,
&\equiv\boxed{\textbf{(D) } 95}\pmod{100}.
 
 
\end{align*}</cmath>
 
\end{align*}</cmath>
 +
from which <math>N\equiv\boxed{\textbf{(D) } 95}\pmod{100}.</math>
 +
 
~MRENTHUSIASM
 
~MRENTHUSIASM
  
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6 & 6 & \tbinom{6}{6}\tbinom{8}{6} & 28 \ [1ex]
 
6 & 6 & \tbinom{6}{6}\tbinom{8}{6} & 28 \ [1ex]
 
\end{array}</cmath>
 
\end{array}</cmath>
We find the total number of such groups: <cmath>70+1+48+336+420+420+1120+160+15+1050+15+48+336+28+28=4095\equiv\boxed{\textbf{(D) } 95}\pmod{100}.</cmath>
+
We find the total number of such groups: <cmath>N=70+1+48+336+420+420+1120+160+15+1050+15+48+336+28+28=4095,</cmath>
 +
from which <math>N\equiv\boxed{\textbf{(D) } 95}\pmod{100}.</math>
 +
 
 
~sugar_rush
 
~sugar_rush
  

Revision as of 11:49, 24 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 1 (Bijection)

Suppose that $t$ tenors and $b$ basses will sing. The requirements are $t\equiv b\pmod{4}$ and $(t,b)\neq(0,0).$

It follows that $b'=8-b$ basses will not sing. Since the ordered pairs $(t,b)$ and the ordered pairs $(t,b')$ have one-to-one correspondence, we consider the ordered pairs $(t,b')$ instead. The requirements become $t\equiv8-b'\pmod{4}$ and $(t,8-b')\neq(0,0),$ which simplify to $t+b'\equiv0\pmod{4}$ and $(t,b')\neq(0,8),$ respectively.

As $t+b'\in\{0,4,6\},$ the total number of such groups is \begin{align*} N&=\binom{14}{0}+\binom{14}{4}+\left[\binom{14}{8}-1\right]+\binom{14}{12} \\ &=\binom{14}{0}+\binom{14}{4}+\left[\binom{14}{6}-1\right]+\binom{14}{2} \\ \end{align*}

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 $4\mid k$, then \[h(i)+h(-1)+h(-i)+h(1)=1+1+1+1=4.\] 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 $4096-1=4095\equiv\boxed{\textbf{(D) } 95}\pmod{100}$.

~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 $\binom{n}{r}=\binom{n}{n-r}:$ \[\begin{array}{c|c|c|c} & & & \\ [-2ex] \textbf{\# of Tenors} & \textbf{\# of Basses} & \textbf{\# of Groups} & \textbf{Rewrite \# of Groups} \\ [0.5ex] \hline\hline  & & & \\ [-2ex] 0 & 8 & \tbinom{6}{0}\tbinom{8}{8} & \\ [1ex] 1 & 1 & \tbinom{6}{1}\tbinom{8}{1} & \tbinom{6}{1}\tbinom{8}{7}\\ [1ex] 2 & 2 & \tbinom{6}{2}\tbinom{8}{2} & \tbinom{6}{2}\tbinom{8}{6}\\ [1ex] 3 & 3 & \tbinom{6}{3}\tbinom{8}{3} & \tbinom{6}{3}\tbinom{8}{5}\\ [1ex] 4 & 4 & \tbinom{6}{4}\tbinom{8}{4} & \\ [1ex] 5 & 5 & \tbinom{6}{5}\tbinom{8}{5} & \tbinom{6}{5}\tbinom{8}{3}\\ [1ex] 6 & 6 & \tbinom{6}{6}\tbinom{8}{6} & \tbinom{6}{6}\tbinom{8}{2}\\ [1ex] \hline & & & \\ [-2ex] 0 & 4 & \tbinom{6}{0}\tbinom{8}{4} & \\ [1ex] 1 & 5 & \tbinom{6}{1}\tbinom{8}{5} & \tbinom{6}{1}\tbinom{8}{3}\\ [1ex] 2 & 6 & \tbinom{6}{2}\tbinom{8}{6} & \tbinom{6}{2}\tbinom{8}{2}\\ [1ex] 3 & 7 & \tbinom{6}{3}\tbinom{8}{7} & \tbinom{6}{3}\tbinom{8}{1}\\ [1ex] 4 & 8 & \tbinom{6}{4}\tbinom{8}{8} & \tbinom{6}{4}\tbinom{8}{0}\\ [1ex] \hline & & & \\ [-2ex] 4 & 0 & \tbinom{6}{4}\tbinom{8}{0} & \tbinom{6}{2}\tbinom{8}{0}\\ [1ex] 5 & 1 & \tbinom{6}{5}\tbinom{8}{1} & \tbinom{6}{1}\tbinom{8}{1}\\ [1ex] 6 & 2 & \tbinom{6}{6}\tbinom{8}{2} & \tbinom{6}{0}\tbinom{8}{2}\\ [1ex] \end{array}\] We apply Vandermonde's Identity to find the total number of such groups: \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, \end{align*} from which $N\equiv\boxed{\textbf{(D) } 95}\pmod{100}.$

~MRENTHUSIASM

Solution 4 (Enumeration)

By casework, we construct the following table: \[\begin{array}{c|c|c|c} & & & \\ [-2ex] \textbf{\# of Tenors} & \textbf{\# of Basses} & \textbf{\# of Groups} & \textbf{Evaluate \# of Groups} \\ [0.5ex] \hline\hline  & & & \\ [-2ex] 0 & 4 & \tbinom{6}{0}\tbinom{8}{4} & 70 \\ [1ex] 0 & 8 & \tbinom{6}{0}\tbinom{8}{8} & 1 \\ [1ex] \hline & & & \\ [-2ex] 1 & 1 & \tbinom{6}{1}\tbinom{8}{1} & 48 \\ [1ex] 1 & 5 & \tbinom{6}{1}\tbinom{8}{5} & 336 \\ [1ex] \hline & & & \\ [-2ex] 2 & 2 & \tbinom{6}{2}\tbinom{8}{2} & 420 \\ [1ex] 2 & 6 & \tbinom{6}{2}\tbinom{8}{6} & 420 \\ [1ex] \hline & & & \\ [-2ex] 3 & 3 & \tbinom{6}{3}\tbinom{8}{3} & 1120 \\ [1ex] 3 & 7 & \tbinom{6}{3}\tbinom{8}{7} & 160 \\ [1ex] \hline & & & \\ [-2ex] 4 & 0 & \tbinom{6}{4}\tbinom{8}{0} & 15 \\ [1ex] 4 & 4 & \tbinom{6}{4}\tbinom{8}{4} & 1050 \\ [1ex] 4 & 8 & \tbinom{6}{4}\tbinom{8}{8} & 15 \\ [1ex] \hline & & & \\ [-2ex] 5 & 1 & \tbinom{6}{5}\tbinom{8}{1} & 48 \\ [1ex] 5 & 5 & \tbinom{6}{5}\tbinom{8}{5} & 336 \\ [1ex] \hline & & & \\ [-2ex] 6 & 2 & \tbinom{6}{6}\tbinom{8}{2} & 28 \\ [1ex] 6 & 6 & \tbinom{6}{6}\tbinom{8}{6} & 28 \\ [1ex] \end{array}\] We find the total number of such groups: \[N=70+1+48+336+420+420+1120+160+15+1050+15+48+336+28+28=4095,\] from which $N\equiv\boxed{\textbf{(D) } 95}\pmod{100}.$

~sugar_rush

~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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