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

(Put all video solutions at the end.)
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~Lopkiloinm
 
~Lopkiloinm
  
−
==Solution 2: Generating Functions==
+
==Solution 2 (Generating Functions)==
  
 
The problem can be done using a roots of unity filter. Let <math>f(x,y)=(1+x)^8(1+y)^6</math>. By expanding the binomials and distributing, <math>f(x,y)</math> is the generating function for different groups of basses and tenors. That is,  
 
The problem can be done using a roots of unity filter. Let <math>f(x,y)=(1+x)^8(1+y)^6</math>. By expanding the binomials and distributing, <math>f(x,y)</math> is the generating function for different groups of basses and tenors. That is,  
Line 34: Line 34:
 
(since <math>g(-1)=0</math> and it can be checked that <math>g(i)=-g(-i)</math>). Hence, the answer is <math>4096-1</math> with the <math>-1</math> for <math>a_{00}</math> which gives <math>\boxed{95}</math>.
 
(since <math>g(-1)=0</math> and it can be checked that <math>g(i)=-g(-i)</math>). Hence, the answer is <math>4096-1</math> with the <math>-1</math> for <math>a_{00}</math> which gives <math>\boxed{95}</math>.
 
~lawliet163
 
~lawliet163
 +
 +
==Solution 3 (Casework and Vandermonde's Identity)==
 +
<cmath>\begin{array}{ccc}
 +
\text{\# of Tenors} & \text{\# of Basses} & \text{\# of ways} \\ \hline\hline
 +
1 & 1 & 1 \\
 +
2 & 2 & 1 \\
 +
3 & 3 & 2 \\
 +
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 \\
 +
4 & 0 & 3 \\
 +
5 & 1 & 4 \\
 +
6 & 2 & 4
 +
\end{array}</cmath>
  
 
==Video Solution by Punxsutawney Phil==
 
==Video Solution by Punxsutawney Phil==

Revision as of 17:39, 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. $\binom{6}{0}\left(\binom{8}{4}+\binom{8}{8}\right)+\binom{6}{1}\left(\binom{8}{1}+\binom{8}{5}\right)+\binom{6}{2}\left(\binom{8}{2}+\binom{8}{6}\right)+\binom{6}{3}\left(\binom{8}{3}+\binom{8}{7}\right)+\\\binom{6}{4}\left(\binom{8}{0}+\binom{8}{4}+\binom{8}{8}\right)+\binom{6}{5}\left(\binom{8}{1}+\binom{8}{5}\right)+\binom{6}{6}\left(\binom{8}{2}+\binom{8}{6}\right) = \boxed{(D) 4095}$

~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)

\[\begin{array}{ccc} \text{\# of Tenors} & \text{\# of Basses} & \text{\# of ways} \\ \hline\hline 1 & 1 & 1 \\ 2 & 2 & 1 \\ 3 & 3 & 2 \\ 4 & 4 & 3 \\ 5 & 5 & 4 \\ 6 & 6 & 5 \\ 0 & 8 & 6 \\ \hline 0 & 4 & 7 \\ 1 & 5 & 8 \\ 2 & 6 & 9 \\ 3 & 7 & 1 \\ 4 & 8 & 2 \\ 4 & 0 & 3 \\ 5 & 1 & 4 \\ 6 & 2 & 4 \end{array}\]

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

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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