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

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==Solution==
 
==Solution==
 
Let <math>c=6-(a+b)</math>, and note that <math>\binom{6}{a + b}=\binom{6}{c}</math>. The problem thus asks for the sum <math>\binom{6}{a} \binom{6}{b} \binom{6}{c}</math> over all <math>a,b,c</math> such that <math>a+b+c=6</math>. 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 <math>\binom{18}{6}=18564</math>. Therefore, the answer is <math>\boxed{564}</math>.
 
Let <math>c=6-(a+b)</math>, and note that <math>\binom{6}{a + b}=\binom{6}{c}</math>. The problem thus asks for the sum <math>\binom{6}{a} \binom{6}{b} \binom{6}{c}</math> over all <math>a,b,c</math> such that <math>a+b+c=6</math>. 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 <math>\binom{18}{6}=18564</math>. Therefore, the answer is <math>\boxed{564}</math>.
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+
-rocketscience
 
==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 22:17, 8 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

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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