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

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

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

Solution 3

Treating $a+b$ as $n$ , this problem asks for:

$\sum_{n=0}^{6}$ ( $\binom{6}{n}$ * $\sum_{m=0}^{n}$ ( $\binom{6}{m}$ * $\binom{6}{n-m}$ ))

But, $\sum_{m=0}^{n} (\binom{6}{m} * \binom{6}{n-m})$ can be seen as 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 $0$ <= $m$ <= $n$ .

Thus, such a procedure is simply $\binom{12}{n}$ .

Therefore, our answer is $\sum_{n=0}^{6} \binom{6}{n} * \binom{12}{n}$ .

- Awsomness2000

@@ Line 48: / Line 48: @@
-==Problem 7==
+== Solution 3 ==
-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>.
-Solution 3:
 Treating <math>a+b</math> as <math>n</math>, this problem asks for:
@@ Line 67: / Line 63: @@
 Therefore, our answer is  <math>\sum_{n=0}^{6} \binom{6}{n} * \binom{12}{n}</math>.
+- Awsomness2000
 ==See Also==
 {{AIME box|year=2017|n=I|num-b=6|num-a=8}}
 {{MAA Notice}}

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