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Difference between revisions of "2017 AIME I Problems/Problem 7"

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<cmath>316+124+124=\boxed{564}</cmath>
 
<cmath>316+124+124=\boxed{564}</cmath>
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==Problem 7==
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For nonnegative integers <math>a</math> and <math>b</math> with  <math>a + b \leq 6</math>, let <math>T(a, b) = \binom{6}{a} \binom{6}{b} \binom{6}{a + b}</math>. Let <math>S</math> denote the sum of all <math>T(a, b)</math>, where  <math>a</math> and <math>b</math> are nonnegative integers with <math>a + b \leq 6</math>. Find the remainder when <math>S</math> is divided by <math>1000</math>.
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Solution 3:
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Treating <math>a+b</math> as <math>n</math>, this problem asks for:
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<math>\sum_{n=0}^{6}</math> (<math>\binom{6}{n}</math> * <math>\sum_{m=0}^{n}</math> (<math>\binom{6}{m}</math> * <math>\binom{6}{n-m}</math>))
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But,
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<math>\sum_{m=0}^{n} (\binom{6}{m} * \binom{6}{n-m})</math>
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can be seen as the following combinatorial argument:
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Choosing <math>n</math> elements from a set of size <math>12</math> is the same as splitting the set into two sets of size <math>6</math> and choosing <math>m</math> elements from one, <math>n-m</math> from the other where <math>0</math> <= <math>m</math> <= <math>n</math> .
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Thus, such a procedure is simply <math>\binom{12}{n}</math>.
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Therefore, our answer is  <math>\sum_{n=0}^{6} \binom{6}{n} * \binom{12}{n}</math>.
  
 
==See Also==
 
==See Also==
 
{{AIME box|year=2017|n=I|num-b=6|num-a=8}}
 
{{AIME box|year=2017|n=I|num-b=6|num-a=8}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 13:33, 12 March 2017

Problem 7

For nonnegative integers $a$ and $b$ with $a + b \leq 6$, let $T(a, b) = \binom{6}{a} \binom{6}{b} \binom{6}{a + b}$. Let $S$ denote the sum of all $T(a, b)$, where $a$ and $b$ are nonnegative integers with $a + b \leq 6$. Find the remainder when $S$ is divided by $1000$.

Solution

Let $c=6-(a+b)$, and note that $\binom{6}{a + b}=\binom{6}{c}$. The problem thus asks for the sum $\binom{6}{a} \binom{6}{b} \binom{6}{c}$ over all $a,b,c$ such that $a+b+c=6$. Consider an array of 18 dots, with 3 columns of 6 dots each. The desired expression counts the total number of ways to select 6 dots by considering each column separately. However, this must be equal to $\binom{18}{6}=18564$. Therefore, the answer is $\boxed{564}$.

-rocketscience


Solution 2 (Major Bash)

Case 1: $a<b$.

Subcase 1: $a=0$ \[\binom{6}{0}\binom{6}{1}\binom{6}{1}=36\] \[\binom{6}{0}\binom{6}{2}\binom{6}{2}=225\] \[\binom{6}{0}\binom{6}{3}\binom{6}{3}=400\] \[\binom{6}{0}\binom{6}{4}\binom{6}{4}=225\] \[\binom{6}{0}\binom{6}{5}\binom{6}{5}=36\] \[\binom{6}{0}\binom{6}{6}\binom{6}{6}=1\] \[36+225+400+225+36+1=923\] Subcase 2: $a=1$ \[\binom{6}{1}\binom{6}{2}\binom{6}{3}=1800 \equiv 800 \pmod {1000}\] \[\binom{6}{1}\binom{6}{3}\binom{6}{4}=1800 \equiv 800 \pmod {1000}\] \[\binom{6}{1}\binom{6}{4}\binom{6}{5}=540\] \[\binom{6}{1}\binom{6}{5}\binom{6}{6}=36\] \[800+800+540+36=2176 \equiv 176 \pmod {1000}\] Subcase 3: $a=2$ \[\binom{6}{2}\binom{6}{3}\binom{6}{5}=1800\equiv800\pmod{1000}\] \[\binom{6}{2}\binom{6}{4}\binom{6}{6}=225\] \[800+225=1025\equiv25\pmod{1000}\]


\[923+176+25=1124\equiv124\pmod{1000}\]

Case 2: $b<a$

By just switching $a$ and $b$ in all of the above cases, we will get all of the cases such that $b>a$ is true. Therefore, this case is also $124\pmod{1000}$

Case 3: $a=b$ \[\binom{6}{0}\binom{6}{0}\binom{6}{0}=1\] \[\binom{6}{1}\binom{6}{1}\binom{6}{2}=540\] \[\binom{6}{2}\binom{6}{2}\binom{6}{4}=3375\equiv375\pmod{1000}\] \[\binom{6}{3}\binom{6}{3}\binom{6}{6}=400\] \[1+540+375+400=1316\equiv316\pmod{1000}\]


\[316+124+124=\boxed{564}\]


Problem 7

For nonnegative integers $a$ and $b$ with $a + b \leq 6$, let $T(a, b) = \binom{6}{a} \binom{6}{b} \binom{6}{a + b}$. Let $S$ denote the sum of all $T(a, b)$, where $a$ and $b$ are nonnegative integers with $a + b \leq 6$. Find the remainder when $S$ is divided by $1000$.


Solution 3:

Treating $a+b$ as $n$, this problem asks for:

$\sum_{n=0}^{6}$ ($\binom{6}{n}$ * $\sum_{m=0}^{n}$ ($\binom{6}{m}$ * $\binom{6}{n-m}$))

But, $\sum_{m=0}^{n} (\binom{6}{m} * \binom{6}{n-m})$ can be seen as the following combinatorial argument:

Choosing $n$ elements from a set of size $12$ is the same as splitting the set into two sets of size $6$ and choosing $m$ elements from one, $n-m$ from the other where $0$ <= $m$ <= $n$ .

Thus, such a procedure is simply $\binom{12}{n}$.

Therefore, our answer is $\sum_{n=0}^{6} \binom{6}{n} * \binom{12}{n}$.

See Also

2017 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 6 		Followed by
Problem 8
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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