Difference between revisions of "2021 AMC 12A Problems/Problem 15"
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<math>\textbf{(A) } 47\qquad\textbf{(B) } 48\qquad\textbf{(C) } 83\qquad\textbf{(D) } 95\qquad\textbf{(E) } 96\qquad</math> | <math>\textbf{(A) } 47\qquad\textbf{(B) } 48\qquad\textbf{(C) } 83\qquad\textbf{(D) } 95\qquad\textbf{(E) } 96\qquad</math> | ||
− | ==Solution 1 ( | + | ==Solution 1 (Bijection)== |
− | < | + | Suppose that <math>t</math> tenors and <math>b</math> basses are selected. 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> basses are 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\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. |
− | 0 | + | |
− | + | As <math>t+b'\in\{0,4,8,12\},</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} \\ | |
− | + | &=1+1001+[3003-1]+91 \\ | |
− | + | &=4095, | |
− | \ | + | \end{align*}</cmath> |
+ | from which <math>N\equiv\boxed{\textbf{(D) } 95}\pmod{100}.</math> | ||
+ | |||
+ | ~MRENTHUSIASM | ||
+ | |||
+ | ==Solution 2 (Vandermonde's Identity)== | ||
+ | Suppose that <math>t</math> tenors and <math>b</math> basses are selected. The requirements are <math>t\equiv b\pmod{4}</math> and <math>(t,b)\neq(0,0).</math> | ||
+ | |||
+ | Note that <math>\binom{6}{t}\binom{8}{b}</math> different groups can be formed by selecting <math>t</math> tenors and <math>b</math> basses. Since <math>t-b\in\{-8,-4,0,4\},</math> we apply casework: | ||
+ | <ol style="margin-left: 1.5em;"> | ||
+ | <li>If <math>t-b=-8,</math> then <math>\binom{6}{0}\binom{8}{8}</math> different group can be formed.</li><p> | ||
+ | <li>If <math>t-b=-4,</math> then <math>\sum_{k=0}^{4}\binom{6}{k}\binom{8}{k+4}</math> different groups can be formed.</li><p> | ||
+ | <li>If <math>t-b=0,</math> then <math>\left[\sum_{k=0}^{6}\binom{6}{k}\binom{8}{k}\right]-1</math> different groups can be formed, recalling that <math>(t,b)\neq(0,0).</math></li><p> | ||
+ | <li>If <math>t-b=4,</math> then <math>\sum_{k=0}^{2}\binom{6}{k+4}\binom{8}{k}</math> different groups can be formed.</li><p> | ||
+ | </ol> | ||
+ | By the combinatorial identity <math>\binom{n}{k}=\binom{n-k}{k}</math> and Vandermonde's Identity <math>\sum_{k=0}^{r}\binom{m}{k}\binom{n}{r-k}=\binom{m+n}{r},</math> we find the total number of such groups: | ||
+ | <cmath>\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*}</cmath> | ||
+ | from which <math>N\equiv\boxed{\textbf{(D) } 95}\pmod{100}.</math> | ||
− | + | ~MRENTHUSIASM | |
− | ==Solution | + | ==Solution 3 (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, <cmath>f(x,y)=\sum_{m=0}^8\sum_{n=0}^6 a_{mn}x^my^n,</cmath> |
− | <cmath> | + | where <math>a_{mn}</math> is the number of groups of <math>m</math> basses and <math>n</math> tenors. What we want to do is sum up all values of <math>a_{mn}</math> for which <math>4\mid m-n</math> except for <math>a_{00}=1</math>. To do this, define a new function <cmath>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.</cmath> |
− | f(x,y)=\sum_{m=0}^8\sum_{n=0}^6 a_{mn}x^my^n | + | Now we just need to sum all coefficients of <math>g(x)</math> for which <math>4\mid m-n</math>. Consider a monomial <math>h(x)=x^k</math>. If <math>4\mid k</math>, then <cmath>h(i)+h(-1)+h(-i)+h(1)=1+1+1+1=4.</cmath> |
− | </cmath> | + | Otherwise, |
− | where <math>a_{mn}</math> is the number of groups of <math>m</math> basses and <math>n</math> tenors. What we want to do is sum up all values of <math>a_{mn}</math> for which <math>4\mid m-n</math> except for <math>a_{00}=1</math>. To do this, define a new function | + | <cmath>h(i)+h(-1)+h(-i)+h(1)=0.</cmath> |
− | <cmath> | ||
− | 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. | ||
− | </cmath> | ||
− | Now we just need to sum all coefficients of <math>g(x)</math> for which <math>4\mid m-n</math>. Consider a monomial <math>h(x)=x^k</math>. If <math>4\mid k</math>, | ||
− | <cmath> | ||
− | h(i)+h(-1)+h(-i)+h(1)=1+1+1+1=4 | ||
− | </cmath> | ||
− | |||
− | <cmath> | ||
− | h(i)+h(-1)+h(-i)+h(1)=0. | ||
− | </cmath> | ||
<math>g(x)</math> is a sum of these monomials so this gives us a method to determine the sum we're looking for: | <math>g(x)</math> is a sum of these monomials so this gives us a method to determine the sum we're looking for: | ||
− | <cmath> | + | <cmath>\frac{g(i)+g(-1)+g(-i)+g(1)}{4}=2^{12}=4096.</cmath> |
− | \frac{g(i)+g(-1)+g(-i)+g(1)}{4}=2^{12}=4096 | + | (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=4095\equiv\boxed{\textbf{(D) } 95}\pmod{100}</math>. |
− | </cmath> | + | |
− | (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 | ||
~lawliet163 | ~lawliet163 | ||
− | ==Solution | + | ==Solution 4 (Enumeration)== |
− | + | Note that <math>\binom{6}{t}\binom{8}{b}</math> different groups can be formed by selecting <math>t</math> tenors and <math>b</math> basses. By casework, we construct the following table: | |
<cmath>\begin{array}{c|c|c|c} | <cmath>\begin{array}{c|c|c|c} | ||
& & & \\ [-2ex] | & & & \\ [-2ex] | ||
− | \textbf{\# of Tenors} & \textbf{\# of Basses} & \textbf{\# of | + | \textbf{\# of Tenors} & \textbf{\# of Basses} & \textbf{\# of Groups} & \textbf{Evaluate \# of Groups} \\ [0.5ex] |
\hline\hline | \hline\hline | ||
& & & \\ [-2ex] | & & & \\ [-2ex] | ||
− | 0 & | + | 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 | \hline | ||
& & & \\ [-2ex] | & & & \\ [-2ex] | ||
− | + | 2 & 2 & \tbinom{6}{2}\tbinom{8}{2} & 420 \\ [1ex] | |
− | + | 2 & 6 & \tbinom{6}{2}\tbinom{8}{6} & 420 \\ [1ex] | |
− | 2 & | ||
− | |||
− | |||
\hline | \hline | ||
& & & \\ [-2ex] | & & & \\ [-2ex] | ||
− | 4 & 0 & \tbinom{6}{4}\tbinom{8}{0} & \tbinom{6}{ | + | 3 & 3 & \tbinom{6}{3}\tbinom{8}{3} & 1120 \\ [1ex] |
− | 5 & 1 & \tbinom{6}{5}\tbinom{8}{1} & \tbinom{6}{ | + | 3 & 7 & \tbinom{6}{3}\tbinom{8}{7} & 160 \\ [1ex] |
− | 6 & 2 & \tbinom{6}{6}\tbinom{8}{2} & \tbinom{6}{ | + | \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}</cmath> | \end{array}</cmath> | ||
− | We | + | 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> |
− | <cmath> | + | from which <math>N\equiv\boxed{\textbf{(D) } 95}\pmod{100}.</math> |
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− | + | Alternatively, since the answer choices have different units digits, it suffices to find the units digit of <math>N</math> only. | |
− | + | ~sugar_rush ~MRENTHUSIASM | |
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− | ~MRENTHUSIASM | ||
==Video Solution by Punxsutawney Phil== | ==Video Solution by Punxsutawney Phil== | ||
Line 114: | Line 103: | ||
https://www.youtube.com/watch?v=AjQARBvdZ20 | https://www.youtube.com/watch?v=AjQARBvdZ20 | ||
− | ==Video Solution by OmegaLearn ( | + | ==Video Solution by OmegaLearn (Using Vandermonde's Identity)== |
https://www.youtube.com/watch?v=mki7xtZLk1I | https://www.youtube.com/watch?v=mki7xtZLk1I | ||
Revision as of 15:19, 29 June 2022
Contents
Problem
A choir director must select a group of singers from among his tenors and basses. The only requirements are that the difference between the number of tenors and basses must be a multiple of , and the group must have at least one singer. Let be the number of different groups that could be selected. What is the remainder when is divided by ?
Solution 1 (Bijection)
Suppose that tenors and basses are selected. The requirements are and
It follows that basses are not selected. Since the ordered pairs and the ordered pairs have one-to-one correspondence, we consider the ordered pairs instead. The requirements become and which simplify to and respectively.
As the total number of such groups is from which
~MRENTHUSIASM
Solution 2 (Vandermonde's Identity)
Suppose that tenors and basses are selected. The requirements are and
Note that different groups can be formed by selecting tenors and basses. Since we apply casework:
- If then different group can be formed.
- If then different groups can be formed.
- If then different groups can be formed, recalling that
- If then different groups can be formed.
By the combinatorial identity and Vandermonde's Identity we find the total number of such groups: from which
~MRENTHUSIASM
Solution 3 (Generating Functions)
The problem can be done using a roots of unity filter. Let . By expanding the binomials and distributing, is the generating function for different groups of basses and tenors. That is, where is the number of groups of basses and tenors. What we want to do is sum up all values of for which except for . To do this, define a new function Now we just need to sum all coefficients of for which . Consider a monomial . If , then Otherwise, is a sum of these monomials so this gives us a method to determine the sum we're looking for: (since and it can be checked that ). Hence, the answer is .
~lawliet163
Solution 4 (Enumeration)
Note that different groups can be formed by selecting tenors and basses. By casework, we construct the following table: We find the total number of such groups: from which
Alternatively, since the answer choices have different units digits, it suffices to find the units digit of only.
~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 (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.