2017 AIME I Problems/Problem 7

Revision as of 13:30, 3 January 2018 by Mathymath (talk | contribs) (Solution 4)

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 $\boxed{564}$.


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: $a=2$ \[\binom{6}{2}\binom{6}{3}\binom{6}{5}=1800\equiv800\pmod{1000}\] \[\binom{6}{2}\binom{6}{4}\binom{6}{6}=225\] \[800+225=1025\equiv25\pmod{1000}\]


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


Solution 4

We begin as in solution 1 to rewrite the sum as $\binom{6}{a} \binom{6}{b} \binom{6}{c}$ over all $a,b,c$ such that $a+b+c=6$. Consider the polynomial $P(x)=\left(\binom{6}{0} + \binom{6}{1}x + \binom{6}{2}x^2 + \binom{6}{3}x^3 + \cdot \cdot \cdot + \binom{6}{6}x^6\right)^3$. We can see the sum we wish to compute is just the coefficient of the $x^6$ term. However $P(x)=((1+x)^6)^3=(1+x)^{18}$. Therefore, the coefficient of the $x^6$ term is just $\binom{18}{6} = 18564$ so the answer is $\boxed{564}$.

- mathymath

See Also

2017 AIME I (ProblemsAnswer KeyResources)
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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