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

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-rocketscience
 
-rocketscience
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==Solution 2 (Major Bash)==
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Case 1: <math>a<b</math>.
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Subcase 1: <math>a=0</math>
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<cmath>\binom{6}{0}\binom{6}{1}\binom{6}{1}=36</cmath>
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<cmath>\binom{6}{0}\binom{6}{2}\binom{6}{2}=225</cmath>
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<cmath>\binom{6}{0}\binom{6}{3}\binom{6}{3}=400</cmath>
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<cmath>\binom{6}{0}\binom{6}{4}\binom{6}{4}=225</cmath>
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<cmath>\binom{6}{0}\binom{6}{5}\binom{6}{5}=36</cmath>
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<cmath>\binom{6}{0}\binom{6}{6}\binom{6}{6}=1</cmath>
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<cmath>36+225+400+225+36+1=923</cmath>
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Subcase 2: <math>a=1</math>
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<cmath>\binom{6}{1}\binom{6}{2}\binom{6}{3}=1800 \equiv 800 \pmod {1000}</cmath>
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<cmath>\binom{6}{1}\binom{6}{3}\binom{6}{4}=1800 \equiv 800 \pmod {1000}</cmath>
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<cmath>\binom{6}{1}\binom{6}{4}\binom{6}{5}=540</cmath>
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<cmath>\binom{6}{1}\binom{6}{5}\binom{6}{6}=36</cmath>
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<cmath>800+800+540+36=2176 \equiv 176 \pmod {1000}</cmath>
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Subcase 3: <math>a=2</math>
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<cmath>\binom{6}{2}\binom{6}{3}\binom{6}{5}=1800\equiv800\pmod{1000}</cmath>
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<cmath>\binom{6}{2}\binom{6}{4}\binom{6}{6}=225</cmath>
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<cmath>800+225=1025\equiv25\pmod{1000}</cmath>
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<cmath>923+176+25=1124\equiv124\pmod{1000}</cmath>
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Case 2: <math>b<a</math>
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By just switching <math>a</math> and <math>b</math> in all of the above cases, we will get all of the cases such that <math>b>a</math> is true. Therefore, this case is also <math>124\pmod{1000}</math>
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Case 3: <math>a=b</math>
 +
<cmath>\binom{6}{0}\binom{6}{0}\binom{6}{0}=1</cmath>
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<cmath>\binom{6}{1}\binom{6}{1}\binom{6}{2}=540</cmath>
 +
<cmath>\binom{6}{2}\binom{6}{2}\binom{6}{4}=3375\equiv375\pmod{1000}</cmath>
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<cmath>\binom{6}{3}\binom{6}{3}\binom{6}{6}=400</cmath>
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<cmath>1+540+375+400=1316\equiv316\pmod{1000}</cmath>
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<cmath>316+124+124=\boxed{564}</cmath>
  
 
==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 21:19, 9 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: $a=b$ \[\binom{6}{0}\binom{6}{0}\binom{6}{0}=1\] \[\binom{6}{1}\binom{6}{1}\binom{6}{2}=540\] \[\binom{6}{2}\binom{6}{2}\binom{6}{4}=3375\equiv375\pmod{1000}\] \[\binom{6}{3}\binom{6}{3}\binom{6}{6}=400\] \[1+540+375+400=1316\equiv316\pmod{1000}\]


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

See Also

2017 AIME I (Problems • Answer Key • Resources)
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Problem 6 		Followed by
Problem 8
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All AIME Problems and Solutions

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