Difference between revisions of "2017 AIME I Problems/Problem 7"
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+ | ==Problem 7== | ||
+ | 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: | ||
+ | |||
+ | <math>\sum_{n=0}^{6}</math> (<math>\binom{6}{n}</math> * <math>\sum_{m=0}^{n}</math> (<math>\binom{6}{m}</math> * <math>\binom{6}{n-m}</math>)) | ||
+ | |||
+ | But, | ||
+ | <math>\sum_{m=0}^{n} (\binom{6}{m} * \binom{6}{n-m})</math> | ||
+ | can be seen as the following combinatorial argument: | ||
+ | |||
+ | Choosing <math>n</math> elements from a set of size <math>12</math> is the same as splitting the set into two sets of size <math>6</math> and choosing <math>m</math> elements from one, <math>n-m</math> from the other where <math>0</math> <= <math>m</math> <= <math>n</math> . | ||
+ | |||
+ | Thus, such a procedure is simply <math>\binom{12}{n}</math>. | ||
+ | |||
+ | Therefore, our answer is <math>\sum_{n=0}^{6} \binom{6}{n} * \binom{12}{n}</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 13:33, 12 March 2017
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
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. However, this must be equal to
. Therefore, the answer is
.
-rocketscience
Solution 2 (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:
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 3:
Treating as
, this problem asks for:
(
*
(
*
))
But,
can be seen as 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
<=
<=
.
Thus, such a procedure is simply .
Therefore, our 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.