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Difference between revisions of "2017 AIME I Problems/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>.
 
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==
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==Solution 1==
 
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>.
  

Revision as of 17:27, 22 December 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 1

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

Treating $a+b$ as $n$, this problem asks for \[\sum_{n=0}^{6} \left[\binom{6}{n}\sum_{m=0}^{n}\left[\binom{6}{m}\binom{6}{n-m}\right]\right].\] But \[\sum_{m=0}^{n} \left[\binom{6}{m} \binom{6}{n-m}\right]\] can be computed through 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 $0\leq m\leq n$. The number of ways to perform such a procedure is simply $\binom{12}{n}$. Therefore, the requested sum is \[\sum_{n=0}^{6} \left[\binom{6}{n} \binom{12}{n}\right] = 18564.\] As such, our answer is $\boxed{564}$.

- Awsomness2000

Solution 3 (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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