Difference between revisions of "2017 AIME I Problems/Problem 7"
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Now just a quick explanation for people who don't fully understand Vandermonde's. Take the first part, <math>\sum_{a+b=c} \binom{6}{a}\binom{6}{b}</math>. Consider <math>2</math> different groups, <math>A</math> and <math>B</math> both of size <math>6</math> people. We wish to chose <math>a</math> peoples from <math>A</math> and <math>b=c-a</math> people from <math>B</math>. In total, we chose <math>c-a+a=c</math> people. We can then draw a bijection towards choosing <math>c</math> people from <math>A\cup B</math>, which has size <math>12</math>. So, it is <math>\binom{12}{c}=\binom{12}{a+b}</math>. Similarly, for <math>\sum_{c=6} \binom{6}{c}\binom{12}{c}</math>, we see that <math>\binom{6}{c}=\binom{6}{6-c}</math>. Now the total is <math>18</math>, and the sum is <math>6</math>. So, we get <math>\binom{18}{6}</math>. See [[committee forming]] for more information ~ firebolt360 | Now just a quick explanation for people who don't fully understand Vandermonde's. Take the first part, <math>\sum_{a+b=c} \binom{6}{a}\binom{6}{b}</math>. Consider <math>2</math> different groups, <math>A</math> and <math>B</math> both of size <math>6</math> people. We wish to chose <math>a</math> peoples from <math>A</math> and <math>b=c-a</math> people from <math>B</math>. In total, we chose <math>c-a+a=c</math> people. We can then draw a bijection towards choosing <math>c</math> people from <math>A\cup B</math>, which has size <math>12</math>. So, it is <math>\binom{12}{c}=\binom{12}{a+b}</math>. Similarly, for <math>\sum_{c=6} \binom{6}{c}\binom{12}{c}</math>, we see that <math>\binom{6}{c}=\binom{6}{6-c}</math>. Now the total is <math>18</math>, and the sum is <math>6</math>. So, we get <math>\binom{18}{6}</math>. See [[committee forming]] for more information ~ firebolt360 | ||
− | == Solution 2 == | + | == Solution 2 (Committee Forming but slightly more bashy)== |
Treating <math>a+b</math> as <math>n</math>, this problem asks for <cmath>\sum_{n=0}^{6} \left[\binom{6}{n}\sum_{m=0}^{n}\left[\binom{6}{m}\binom{6}{n-m}\right]\right].</cmath> But <cmath>\sum_{m=0}^{n} \left[\binom{6}{m} \binom{6}{n-m}\right]</cmath> | Treating <math>a+b</math> as <math>n</math>, this problem asks for <cmath>\sum_{n=0}^{6} \left[\binom{6}{n}\sum_{m=0}^{n}\left[\binom{6}{m}\binom{6}{n-m}\right]\right].</cmath> But <cmath>\sum_{m=0}^{n} \left[\binom{6}{m} \binom{6}{n-m}\right]</cmath> |
Revision as of 20:21, 24 February 2021
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
.
Major Note
Most solutions use committee forming (except for the bash solution). To understand more about the techniques used, visit the committee forming page for more information.
Solution 1 (Committee Forming)
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 1 but different (Committee Forming)
Alternatively, one can note that we can consider groups where is constant, say
. Fix any value of
. Then the sum of all of the values of
such that
is
which by Vandermonde's is
. Remember, that expression is the resulting sum for a fixed
. So, for
, we want
. This is (by Vandermonde's or committee forming)
~ firebolt360
Note
Now just a quick explanation for people who don't fully understand Vandermonde's. Take the first part, . Consider
different groups,
and
both of size
people. We wish to chose
peoples from
and
people from
. In total, we chose
people. We can then draw a bijection towards choosing
people from
, which has size
. So, it is
. Similarly, for
, we see that
. Now the total is
, and the sum is
. So, we get
. See committee forming for more information ~ firebolt360
Solution 2 (Committee Forming but slightly more bashy)
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
.
Solution 6
Since , we can rewrite
as
. 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
democrats, then pick
republicans, provided that
. Then we can pick the remaining
people from the independents. But this is just
, so the sum of all
is equal to the number of ways to choose this committee.
On the other hand, we can simply pick any 6 people from the
total politicians in the group. Clearly, there are
ways to do this. So the desired quantity is equal to
. We can then compute (routinely) the last 3 digits of
as
.
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 |
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