Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m (→‎Solution 4 (Enumeration))
(→‎Solution 4 (Enumeration))
Line 96: Line 96:
  
 
~MRENTHUSIASM
 
~MRENTHUSIASM
 +
 +
*Small Note: notice all of the answers have different unit's digits, so it suffices to find the units digit, which is much simpler.
  
 
==Video Solution by Punxsutawney Phil==
 
==Video Solution by Punxsutawney Phil==

Revision as of 21:07, 28 June 2022

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 are selected. The requirements are $t\equiv b\pmod{4}$ and $(t,b)\neq(0,0).$

It follows that $b'=8-b$ basses are not selected. 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 $t+b'\in\{0,4,8,12\},$ 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} \\ &=1+1001+[3003-1]+91 \\ &=4095, \end{align*} from which $N\equiv\boxed{\textbf{(D) } 95}\pmod{100}.$

~MRENTHUSIASM

Solution 2 (Vandermonde's Identity)

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

Note that $\binom{6}{t}\binom{8}{b}$ different groups can be formed by selecting $t$ tenors and $b$ basses. Since $t-b\in\{-8,-4,0,4\},$ we apply casework:

  1. If $t-b=-8,$ then $\binom{6}{0}\binom{8}{8}$ different group can be formed.

  2. If $t-b=-4,$ then $\sum_{k=0}^{4}\binom{6}{k}\binom{8}{k+4}$ different groups can be formed.

  3. If $t-b=0,$ then $\left[\sum_{k=0}^{6}\binom{6}{k}\binom{8}{k}\right]-1$ different groups can be formed, recalling that $(t,b)\neq(0,0).$

  4. If $t-b=4,$ then $\sum_{k=0}^{2}\binom{6}{k+4}\binom{8}{k}$ different groups can be formed.

By the combinatorial identity $\binom{n}{k}=\binom{n-k}{k}$ and Vandermonde's Identity $\sum_{k=0}^{r}\binom{m}{k}\binom{n}{r-k}=\binom{m+n}{r},$ we find the total number of such groups: \begin{align*} N&=\binom{6}{0}\binom{8}{8}+\left[\sum_{k=0}^{4}\binom{6}{k}\binom{8}{k+4}\right]+\left[\left[\sum_{k=0}^{6}\binom{6}{k}\binom{8}{k}\right]-1\right]+\left[\sum_{k=0}^{2}\binom{6}{k+4}\binom{8}{k}\right] \\ &=\binom{6}{0}\binom{8}{0}+\left[\sum_{k=0}^{4}\binom{6}{k}\binom{8}{4-k}\right]+\left[\left[\sum_{k=0}^{6}\binom{6}{6-k}\binom{8}{k}\right]-1\right]+\left[\sum_{k=0}^{2}\binom{6}{2-k}\binom{8}{k}\right] \\ &=\binom{14}{0}+\binom{14}{4}+\left[\binom{14}{6}-1\right]+\binom{14}{2} \\ &=1+1001+[3003-1]+91 \\ &=4095, \end{align*} from which $N\equiv\boxed{\textbf{(D) } 95}\pmod{100}.$

~MRENTHUSIASM

Solution 3 (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 4 (Enumeration)

Note that $\binom{6}{t}\binom{8}{b}$ different groups can be formed by selecting $t$ tenors and $b$ basses. 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

  • Small Note: notice all of the answers have different unit's digits, so it suffices to find the units digit, which is much simpler.

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_12A_Problems/Problem_15&oldid=175443"