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

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

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

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 $c=6-(a+b)$ . Then $\binom{6}{a+b}=\binom{6}{c}$ , and $a+b+c=6$ . The problem thus asks for $\sum_{a+b+c=6}\binom{6}{a}\binom{6}{b}\binom{6}{c} \pmod {1000}.$ Suppose we have $6$ red balls, $6$ green balls, and $6$ blue balls lined up in a row, and we want to choose $6$ balls from this set of $18$ balls by considering each color separately. Over all possible selections of $6$ balls from this set, there are always a nonnegative number of balls in each color group. The answer is $\binom{18}{6} \pmod {1000}=18\boxed{564}$ .

Solution 6

Since $\binom{6}{n}=\binom{6}{6-n}$ , we can rewrite $T(a,b)$ as $\binom{6}{a}\binom{6}{b}\binom{6}{6-(a+b)}$ . Consider the number of ways to choose a committee of 6 people from a group of 6 democrats, 6 republicans, and 6 independents. We can first pick $a$ democrats, then pick $b$ republicans, provided that $a+b \leq 6$ . Then we can pick the remaining $6-(a+b)$ people from the independents. But this is just $T(a,b)$ , so the sum of all $T(a,b)$ is equal to the number of ways to choose this committee. On the other hand, we can simply pick any 6 people from the $6+6+6=18$ total politicians in the group. Clearly, there are $\binom{18}{6}$ ways to do this. So the desired quantity is equal to $\binom{18}{6}$ . We can then compute (routinely) the last 3 digits of $\binom{18}{6}$ as $\boxed{564}$ .

@@ Line 63: / Line 63: @@
 ==Solution 5==
 Let <math>c=6-(a+b)</math>. Then <math>\binom{6}{a+b}=\binom{6}{c}</math>, and <math>a+b+c=6</math>. The problem thus asks for <cmath>\sum_{a+b+c=6}\binom{6}{a}\binom{6}{b}\binom{6}{c} \pmod {1000}.</cmath> Suppose we have <math>6</math> red balls, <math>6</math> green balls, and <math>6</math> blue balls lined up in a row, and we want to choose <math>6</math> balls from this set of <math>18</math> balls by considering each color separately. Over all possible selections of <math>6</math> balls from this set, there are always a nonnegative number of balls in each color group. The answer is <math>\binom{18}{6} \pmod {1000}=18\boxed{564}</math>.
-Solution and <math>\LaTeX</math> by ~IceMatrix2
 ==Solution 6==

2017 AIME I (Problems • Answer Key • Resources)
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