2017 AIME I Problems/Problem 7

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

$316+124+124=\boxed{564}$

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

Solution 5

Let's visualize the problem as this: $6$ boys: (Let's call them Eddie, Edmund, Luke, Kareem, Sharjeel, and Bob) $6$ girls: (Let's call them Zuzu, Edmund's imaginary girlfriend, Luke's imaginary girlfriend, Kareem's imaginary girlfriend, Sharjeel's imaginary girlfriend, and Billie (transgender girl)) $6$ animals/pets: (Let's call them Woof, Meow, Bark, Wolf2, Meow2, Bark2) Now, we need to choose $a$ boys, $b$ girls, and $a+b$ animals to be qualified to go to the upcoming prom. Note that choosing $a+b$ animals out of the $6$ total animals to go to the prom is equivalent to choosing $6-a-b$ animals out of the $6$ total animals to NOT go to the prom. Thus, we can rewrite the sum as $\binom{6}{a} \binom{6}{b} \binom{6}{c}$ over all $a,b,c$ such that $a+b+c=6$ . (For instance, if $a = b = 3$ and $6 - a - b = c = 0$ , we have $\dbinom{6}{3}\dbinom{6}{3}\dbinom{6}{0}$ ways, in which one of the possible ways is the set being $\{\text{Eddie}, \text{Zuzu}, \text{Luke}, \text{Luke's imaginary girlfriend}, \text{Edmund}, \text{Edmund's imaginary girlfriend}, \text{Woof}, \text{Woof2}, \text{Meow}, \text{Meow2}, \text{Bark}, \text{Bark2} \}$ . ) It is easy to see that the aforementioned sum is simply the number of ways to choose $6$ total entities from $18$ total entities, or $\dbinom{18}{6} = 18564$ , in which the last three digits are $\boxed{564}$ .

2017 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 6	Followed by Problem 8
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