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

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<math>(1) \ |b-t|\equiv\pm1\pmod{4}</math>
 
<math>(1) \ |b-t|\equiv\pm1\pmod{4}</math>
  
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Clearly,  
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Clearly, the mapping is satisfied. For each group from the <math>12,</math> we can obtain a desired group from the <math>14</math> by adding one tenor or one bass accordingly.
  
 
<math>(2) \ |b-t|\equiv0\pmod{4}</math>
 
<math>(2) \ |b-t|\equiv0\pmod{4}</math>
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Since <math>\binom7b=\binom{7}{7-b},</math> we can select <math>7-b</math> basses instead of <math>b</math> basses, without changing the number of groups. Therefore, we have <math>|(7-b)-t|=|7-(b+t)|.</math> Since <math>b\equiv t\pmod{4},</math> we conclude that <math>|7-(b+t)|\equiv 1\text{ or }3\pmod{4},</math> and the mapping is satisfied by case <math>(1).</math>
  
 
<math>(2) \ |b-t|\equiv2\pmod{4}</math>
 
<math>(2) \ |b-t|\equiv2\pmod{4}</math>

Revision as of 01:26, 17 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)

\[\begin{array}{cccc} \text{\# of Tenors} & \text{\# of Basses} & \text{\# of Ways} & \text{Rewrite \# of Ways} \\ [1ex] \hline\hline & & & \\ [-1ex] 0 & 8 & \binom{6}{0}\binom{8}{8} & \\ [1ex] 1 & 1 & \binom{6}{1}\binom{8}{1} & \binom{6}{1}\binom{8}{7}\\ [1ex] 2 & 2 & \binom{6}{2}\binom{8}{2} & \binom{6}{2}\binom{8}{6}\\ [1ex] 3 & 3 & \binom{6}{3}\binom{8}{3} & \binom{6}{3}\binom{8}{5}\\ [1ex] 4 & 4 & \binom{6}{4}\binom{8}{4} & \\ [1ex] 5 & 5 & \binom{6}{5}\binom{8}{5} & \binom{6}{5}\binom{8}{3}\\ [1ex] 6 & 6 & \binom{6}{6}\binom{8}{6} & \binom{6}{6}\binom{8}{2}\\ [1ex] \hline & & & \\ [-1ex] 0 & 4 & \binom{6}{0}\binom{8}{4} & \\ [1ex] 1 & 5 & \binom{6}{1}\binom{8}{5} & \binom{6}{1}\binom{8}{3}\\ [1ex] 2 & 6 & \binom{6}{2}\binom{8}{6} & \binom{6}{2}\binom{8}{2}\\ [1ex] 3 & 7 & \binom{6}{3}\binom{8}{7} & \binom{6}{3}\binom{8}{1}\\ [1ex] 4 & 8 & \binom{6}{4}\binom{8}{8} & \binom{6}{4}\binom{8}{0}\\ [1ex] \hline & & & \\ [-1ex] 4 & 0 & \binom{6}{4}\binom{8}{0} & \binom{6}{2}\binom{8}{0}\\ [1ex] 5 & 1 & \binom{6}{5}\binom{8}{1} & \binom{6}{1}\binom{8}{1}\\ [1ex] 6 & 2 & \binom{6}{6}\binom{8}{2} & \binom{6}{0}\binom{8}{2} \end{array}\] We will use the Vandermonde's Identity to find the requested sum: \begin{align*} \left[\sum_{k=0}^{6}\binom{6}{k}\binom{8}{8-k}\right]+\left[\sum_{k=0}^{4}\binom{6}{k}\binom{8}{4-k}\right]+\left[\sum_{k=0}^{2}\binom{6}{k}\binom{8}{2-k}\right]&=\binom{14}{8}+\binom{14}{4}+\binom{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

Solution 3 (Combinatoric Argument)

We claim that if the empty group is allowed, then there are $\mathbf{2^{12}=4096}$ ways to choose the singers satisfying the requirements.

First, we set one tenor and one bass aside. We argue that each group from the $12$ remaining singers (of any size, including $0$) corresponds to exactly one desired group.

The $12$ remaining singers can form \[2^{12}=\sum_{\substack{t=0 \\ b=0}}^{\substack{t=5 \\ b=7}}\binom5t\binom7b\] groups. The left side counts directly, while the right side uses casework (selecting $t$ tenors and $b$ basses for each group). Now, we map each group from the remaining $12$ singers to a group from the original $14$ singers.

By casework:

$(1) \ |b-t|\equiv\pm1\pmod{4}$

Clearly, the mapping is satisfied. For each group from the $12,$ we can obtain a desired group from the $14$ by adding one tenor or one bass accordingly.

$(2) \ |b-t|\equiv0\pmod{4}$

Since $\binom7b=\binom{7}{7-b},$ we can select $7-b$ basses instead of $b$ basses, without changing the number of groups. Therefore, we have $|(7-b)-t|=|7-(b+t)|.$ Since $b\equiv t\pmod{4},$ we conclude that $|7-(b+t)|\equiv 1\text{ or }3\pmod{4},$ and the mapping is satisfied by case $(1).$

$(2) \ |b-t|\equiv2\pmod{4}$

~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 (Problems • Answer Key • Resources)
Preceded by
Problem 14 		Followed by
Problem 16
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All AMC 12 Problems and Solutions

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