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

m (→‎Solution 3 (Casework and Vandermonde's Identity))
m (→‎Solution 4 (Combinatoric Argument): Italicized the bolded text as well.)
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==Solution 4 (Combinatoric Argument)==
 
==Solution 4 (Combinatoric Argument)==
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<b>We claim that if the empty group is allowed, then there are <math>\mathbf{2^{12}}</math> ways to choose the singers satisfying the requirements.</b>
+
<i><b>We claim that if the empty group is allowed, then there are <math>\mathbf{2^{12}}</math> ways to choose the singers satisfying the requirements.</b></i>
  
 
First, we set one tenor and one bass aside. We argue that each group from the <math>12</math> remaining singers (of any size, including <math>0</math>) corresponds to exactly one desired group from the original <math>14</math> singers.
 
First, we set one tenor and one bass aside. We argue that each group from the <math>12</math> remaining singers (of any size, including <math>0</math>) corresponds to exactly one desired group from the original <math>14</math> singers.
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By the same reasoning as Case <math>(2),</math> we select <math>7-b</math> basses instead of <math>b</math> basses. The absolute difference also is <math>|7-(b+t)|.</math> Since <math>b-t</math> is even, it follows that <math>b+t</math> is also even, and <math>|7-(b+t)|\equiv 1\text{ or }3\pmod{4}.</math> The mapping is satisfied by case <math>(1).</math>
 
By the same reasoning as Case <math>(2),</math> we select <math>7-b</math> basses instead of <math>b</math> basses. The absolute difference also is <math>|7-(b+t)|.</math> Since <math>b-t</math> is even, it follows that <math>b+t</math> is also even, and <math>|7-(b+t)|\equiv 1\text{ or }3\pmod{4}.</math> The mapping is satisfied by case <math>(1).</math>
  
−
<b>Now, the proof of the bolded claim is complete.</b>
+
<i><b>Now, the proof of the bolded claim is complete.</b></i>
  
 
Therefore, excluding the empty group gives <math>N=2^{12}-1=4095\equiv\boxed{\textbf{(D) } 95}\pmod{100}.</math>
 
Therefore, excluding the empty group gives <math>N=2^{12}-1=4095\equiv\boxed{\textbf{(D) } 95}\pmod{100}.</math>

Revision as of 14:22, 29 March 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 without words

\[\begin{array}{c|c|c} \text{Tenors} & \text{Basses} & \text{Groups} \\ \hline 0 & 4, 8 & \binom{6}{0}\left[\binom{8}{4}+\binom{8}{8}\right]=70+1=\fbox{71} \\ 1 & 1, 5 & \binom{6}{1}\left[\binom{8}{1}+\binom{8}{5}\right]=6(8+56)=6(64)=3\fbox{84} \\ 2 & 2, 6 & \binom{6}{2}\left[\binom{8}{2}+\binom{8}{6}\right]=15(28+28)=15(56)=8\fbox{40} \\ 3 & 3, 7 & \binom{6}{3}\left[\binom{8}{3}+\binom{8}{7}\right]=20(56+8)=20(64)=12\fbox{80} \\ 4 & 0, 4, 8 & \binom{6}{4}\left[\binom{8}{0}+\binom{8}{4}+\binom{8}{8}\right]=15(1+70+1)=15(72)=10\fbox{80} \\ 5 & 1, 5 & \binom{6}{5}\left[\binom{8}{1}+\binom{8}{5}\right]=6(8+56)=6(64)=3\fbox{84} \\ 6 & 2, 6 & \binom{6}{6}\left[\binom{8}{2}+\binom{8}{6}\right]=1(28+28)=\fbox{56} \end{array}\]

$71+384+840+1280+1080+384+56=40\boxed{\textbf{(D)} ~95}$

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)

\[\begin{array}{c|c|c|c} & & & \\ [-2ex] \textbf{\# of Tenors} & \textbf{\# of Basses} & \textbf{\# of Ways} & \textbf{Rewrite \# of Ways} \\ [0.5ex] \hline\hline & & & \\ [-2ex] 0 & 8 & \binom{6}{0}\binom{8}{8} & \\ [1ex] 1 & 1 & \binom{6}{1}\binom{8}{1} & \binom{6}{1}\binom{8}{7}\\ [1ex] 2 & 2 & \binom{6}{2}\binom{8}{2} & \binom{6}{2}\binom{8}{6}\\ [1ex] 3 & 3 & \binom{6}{3}\binom{8}{3} & \binom{6}{3}\binom{8}{5}\\ [1ex] 4 & 4 & \binom{6}{4}\binom{8}{4} & \\ [1ex] 5 & 5 & \binom{6}{5}\binom{8}{5} & \binom{6}{5}\binom{8}{3}\\ [1ex] 6 & 6 & \binom{6}{6}\binom{8}{6} & \binom{6}{6}\binom{8}{2}\\ [1ex] \hline & & & \\ [-2ex] 0 & 4 & \binom{6}{0}\binom{8}{4} & \\ [1ex] 1 & 5 & \binom{6}{1}\binom{8}{5} & \binom{6}{1}\binom{8}{3}\\ [1ex] 2 & 6 & \binom{6}{2}\binom{8}{6} & \binom{6}{2}\binom{8}{2}\\ [1ex] 3 & 7 & \binom{6}{3}\binom{8}{7} & \binom{6}{3}\binom{8}{1}\\ [1ex] 4 & 8 & \binom{6}{4}\binom{8}{8} & \binom{6}{4}\binom{8}{0}\\ [1ex] \hline & & & \\ [-2ex] 4 & 0 & \binom{6}{4}\binom{8}{0} & \binom{6}{2}\binom{8}{0}\\ [1ex] 5 & 1 & \binom{6}{5}\binom{8}{1} & \binom{6}{1}\binom{8}{1}\\ [1ex] 6 & 2 & \binom{6}{6}\binom{8}{2} & \binom{6}{0}\binom{8}{2}\\ [1ex] \end{array}\] 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 4 (Combinatoric Argument)

We claim that if the empty group is allowed, then there are $\mathbf{2^{12}}$ 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 from the original $14$ singers.

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 $12$ to a group from the $14.$

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 the absolute difference $|(7-b)-t|=|7-(b+t)|.$ 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 $(1).$

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

By the same reasoning as Case $(2),$ we select $7-b$ basses instead of $b$ basses. The absolute difference also is $|7-(b+t)|.$ Since $b-t$ is even, it follows that $b+t$ is also even, and $|7-(b+t)|\equiv 1\text{ or }3\pmod{4}.$ The mapping is satisfied by case $(1).$

Now, the proof of the bolded claim is complete.

Therefore, excluding the empty group gives $N=2^{12}-1=4095\equiv\boxed{\textbf{(D) } 95}\pmod{100}.$

~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
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

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