Difference between revisions of "2017 AIME I Problems/Problem 7"
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+ | 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>. | ||
==See Also== | ==See Also== | ||
{{AIME box|year=2017|n=I|num-b=6|num-a=8}} | {{AIME box|year=2017|n=I|num-b=6|num-a=8}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 17:03, 29 September 2020
Contents
Problem 7
For nonnegative integers and
with
, let
. Let
denote the sum of all
, where
and
are nonnegative integers with
. Find the remainder when
is divided by
.
Solution 1
Let , and note that
. The problem thus asks for the sum
over all
such that
. 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
. Therefore, the answer is
.
-rocketscience
Solution 2
Treating as
, this problem asks for
But
can be computed through the following combinatorial argument. Choosing
elements from a set of size
is the same as splitting the set into two sets of size
and choosing
elements from one,
from the other where
. The number of ways to perform such a procedure is simply
. Therefore, the requested sum is
As such, our answer is
.
- Awsomness2000
Solution 3 (Major Major Bash)
Case 1: .
Subcase 1:
Subcase 2:
Subcase 3:
Case 2:
By just switching and
in all of the above cases, we will get all of the cases such that
is true. Therefore, this case is also
Case 3:
Solution 4
We begin as in solution 1 to rewrite the sum as over all
such that
.
Consider the polynomial
.
We can see the sum we wish to compute is just the coefficient of the
term. However
. Therefore, the coefficient of the
term is just
so the answer is
.
- mathymath
Solution 5
Let . Then
, and
. The problem thus asks for
Suppose we have
red balls,
green balls, and
blue balls lined up in a row, and we want to choose
balls from this set of
balls by considering each color separately. Over all possible selections of
balls from this set, there are always a nonnegative number of balls in each color group. The answer is
.
See Also
2017 AIME I (Problems • Answer Key • Resources) | ||
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.