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 (Without Words)==
+
==Solution 1 (Bijection)==
<cmath>\begin{array}{c|c|c}
+
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>
\text{Tenors} & \text{Basses} & \text{Groups} \\  
+
 
\hline
+
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 & 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} \\
+
As <math>t+b'\in\{0,4,8,12\},</math> the total number of such groups is
2 & 2, 6 & \binom{6}{2}\left[\binom{8}{2}+\binom{8}{6}\right]=15(28+28)=15(56)=8\fbox{40} \\
+
<cmath>\begin{align*}
3 & 3, 7 & \binom{6}{3}\left[\binom{8}{3}+\binom{8}{7}\right]=20(56+8)=20(64)=12\fbox{80} \\
+
N&=\binom{14}{0}+\binom{14}{4}+\left[\binom{14}{8}-1\right]+\binom{14}{12} \\
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} \\
+
&=\binom{14}{0}+\binom{14}{4}+\left[\binom{14}{6}-1\right]+\binom{14}{2} \\
5 & 1, 5 & \binom{6}{5}\left[\binom{8}{1}+\binom{8}{5}\right]=6(8+56)=6(64)=3\fbox{84} \\
+
&=1+1001+[3003-1]+91 \\
6 & 2, 6 & \binom{6}{6}\left[\binom{8}{2}+\binom{8}{6}\right]=1(28+28)=\fbox{56}  
+
&=4095,
\end{array}</cmath>
+
\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>
  
<math>71+384+840+1280+1080+384+56=40\boxed{\textbf{(D)} ~95}</math>
+
~MRENTHUSIASM
  
==Solution 2 (Generating Functions)==
+
==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>
 
otherwise,  
 
<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</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)==
+
==Solution 4 (Enumeration)==
By casework, we construct the following table. In the last column, we rewrite some of the combinations using the identity <math>\binom{n}{r}=\binom{n}{n-r}:</math>
+
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 Ways} & \textbf{Rewrite \# of Ways} \\ [0.5ex]
+
\textbf{\# of Tenors} & \textbf{\# of Basses} & \textbf{\# of Groups} & \textbf{Evaluate \# of Groups} \\ [0.5ex]
 
\hline\hline  
 
\hline\hline  
 
& & & \\ [-2ex]
 
& & & \\ [-2ex]
0 & 8 & \tbinom{6}{0}\tbinom{8}{8} & \\ [1ex]
+
0 & 4 & \tbinom{6}{0}\tbinom{8}{4} & 70 \\ [1ex]
1 & 1 & \tbinom{6}{1}\tbinom{8}{1} & \tbinom{6}{1}\tbinom{8}{7}\\ [1ex]
+
0 & 8 & \tbinom{6}{0}\tbinom{8}{8} & 1 \\ [1ex]
2 & 2 & \tbinom{6}{2}\tbinom{8}{2} & \tbinom{6}{2}\tbinom{8}{6}\\ [1ex]
+
\hline
3 & 3 & \tbinom{6}{3}\tbinom{8}{3} & \tbinom{6}{3}\tbinom{8}{5}\\ [1ex]
+
& & & \\ [-2ex]
4 & 4 & \tbinom{6}{4}\tbinom{8}{4} & \\ [1ex]
+
1 & 1 & \tbinom{6}{1}\tbinom{8}{1} & 48 \\ [1ex]
5 & 5 & \tbinom{6}{5}\tbinom{8}{5} & \tbinom{6}{5}\tbinom{8}{3}\\ [1ex]
+
1 & 5 & \tbinom{6}{1}\tbinom{8}{5} & 336 \\ [1ex]
6 & 6 & \tbinom{6}{6}\tbinom{8}{6} & \tbinom{6}{6}\tbinom{8}{2}\\ [1ex]
+
\hline
 +
& & & \\ [-2ex]
 +
2 & 2 & \tbinom{6}{2}\tbinom{8}{2} & 420 \\ [1ex]
 +
2 & 6 & \tbinom{6}{2}\tbinom{8}{6} & 420 \\ [1ex]
 
\hline
 
\hline
 
& & & \\ [-2ex]
 
& & & \\ [-2ex]
0 & 4 & \tbinom{6}{0}\tbinom{8}{4} & \\ [1ex]
+
3 & 3 & \tbinom{6}{3}\tbinom{8}{3} & 1120 \\ [1ex]
1 & 5 & \tbinom{6}{1}\tbinom{8}{5} & \tbinom{6}{1}\tbinom{8}{3}\\ [1ex]
+
3 & 7 & \tbinom{6}{3}\tbinom{8}{7} & 160 \\ [1ex]
2 & 6 & \tbinom{6}{2}\tbinom{8}{6} & \tbinom{6}{2}\tbinom{8}{2}\\ [1ex]
 
3 & 7 & \tbinom{6}{3}\tbinom{8}{7} & \tbinom{6}{3}\tbinom{8}{1}\\ [1ex]
 
4 & 8 & \tbinom{6}{4}\tbinom{8}{8} & \tbinom{6}{4}\tbinom{8}{0}\\ [1ex]
 
 
\hline
 
\hline
 
& & & \\ [-2ex]
 
& & & \\ [-2ex]
4 & 0 & \tbinom{6}{4}\tbinom{8}{0} & \tbinom{6}{2}\tbinom{8}{0}\\ [1ex]
+
4 & 0 & \tbinom{6}{4}\tbinom{8}{0} & 15 \\ [1ex]
5 & 1 & \tbinom{6}{5}\tbinom{8}{1} & \tbinom{6}{1}\tbinom{8}{1}\\ [1ex]
+
4 & 4 & \tbinom{6}{4}\tbinom{8}{4} & 1050 \\ [1ex]
6 & 2 & \tbinom{6}{6}\tbinom{8}{2} & \tbinom{6}{0}\tbinom{8}{2}\\ [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 apply Vandermonde's Identity to find the requested sum:
+
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>\begin{align*}
+
from which <math>N\equiv\boxed{\textbf{(D) } 95}\pmod{100}.</math>
N&=\underbrace{\left[\sum_{k=0}^{6}\binom{6}{k}\binom{8}{8-k}\right]}_{\tbinom{14}{8}}+\underbrace{\left[\sum_{k=0}^{4}\binom{6}{k}\binom{8}{4-k}\right]}_{\tbinom{14}{4}}+\underbrace{\left[\sum_{k=0}^{2}\binom{6}{k}\binom{8}{2-k}\right]}_{\tbinom{14}{2}} \\
 
&=\binom{14}{6}+\binom{14}{4}+\binom{14}{2} \\
 
&=3003+1001+91 \\
 
&=4095 \\
 
&\equiv\boxed{\textbf{(D) } 95}\pmod{100}.
 
\end{align*}</cmath>
 
 
 
~MRENTHUSIASM
 
 
 
==Solution 4 (Combinatoric Argument)==
 
<i><b>We claim that if the empty group is allowed, then there are <math>\mathbf{2^{12}}</math> valid ways to choose the singers.</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 valid group from the original <math>14</math> singers.
 
 
 
The <math>12</math> remaining singers can form <cmath>2^{12}=\left[\sum_{t=0}^{5}\binom5t\right]\cdot\left[\sum_{b=0}^{7}\binom7b\right]</cmath> groups. The left side counts directly, while the right side uses casework (selecting <math>t</math> tenors and <math>b</math> basses for each group). Now, we map each group from the <math>12</math> to one valid group from the original <math>14.</math>
 
 
 
By casework:
 
 
 
<ol style="margin-left: 1.5em;">
 
  <li><math>|b-t|\equiv1\text{ or }3\pmod{4}</math></li><p>
 
Clearly, the mapping is satisfied. For each group from the <math>12,</math> we can obtain one valid group from the original <math>14</math> by adding one tenor or one bass accordingly.
 
  <li><math>|b-t|\equiv0\pmod{4}</math></li><p>
 
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 the absolute difference <math>\left|(7-b)-t\right|=\left|7-(b+t)\right|.</math> Since <math>b\equiv t\pmod{4},</math> we conclude that <math>b+t</math> must be even. It follows that <math>\left|(7-b)-t\right|=\left|7-(b+t)\right|\equiv 1\text{ or }3\pmod{4}.</math> This mapping is satisfied by Case 1.
 
  <li><math>|b-t|\equiv2\pmod{4}</math></li><p>
 
By the same reasoning as Case 2, we select <math>7-b</math> basses instead of <math>b</math> basses. The absolute difference is also <math>\left|7-(b+t)\right|.</math> Since <math>b-t</math> is even, we conclude that <math>b+t</math> must also be even. It follows that <math>\left|7-(b+t)\right|\equiv 1\text{ or }3\pmod{4}.</math> This mapping is satisfied by Case 1.
 
</ol>
 
 
 
<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>
+
~sugar_rush
  
 
~MRENTHUSIASM
 
~MRENTHUSIASM
Line 112: Line 103:
 
https://www.youtube.com/watch?v=AjQARBvdZ20
 
https://www.youtube.com/watch?v=AjQARBvdZ20
  
==Video Solution by OmegaLearn (using Vandermonde's Identity)==
+
==Video Solution by OmegaLearn (Using Vandermonde's Identity)==
 
https://www.youtube.com/watch?v=mki7xtZLk1I
 
https://www.youtube.com/watch?v=mki7xtZLk1I
  

Latest revision as of 18:11, 24 August 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 (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)

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

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 (ProblemsAnswer KeyResources)
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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