Difference between revisions of "2017 AIME I Problems/Problem 7"
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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>. | 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 1== | + | ==Major Note== |
− | Let <math>c=6-(a+b)</math>, and note that <math>\binom{6}{a + b}=\binom{6}{c}</math>. The problem thus asks for the sum <math>\binom{6}{a} \binom{6}{b} \binom{6}{c}</math> over all <math>a,b,c</math> such that <math>a+b+c=6</math>. 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 | + | 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 <math>c=6-(a+b)</math>, and note that <math>\binom{6}{a + b}=\binom{6}{c}</math>. The problem thus asks for the sum <math>\binom{6}{a} \binom{6}{b} \binom{6}{c}</math> over all <math>a,b,c</math> such that <math>a+b+c=6</math>. 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 <math>\binom{18}{6}=18564</math>. Therefore, the answer is <math>\boxed{564}</math>. | ||
-rocketscience | -rocketscience | ||
− | == Solution 2 == | + | ==Solution 1 but different (Committee Forming)== |
+ | Alternatively, one can note that we can consider groups where <math>a+b</math> is constant, say <math>c</math>. Fix any value of <math>c</math>. Then the sum of all of the values of <math>T(a,b)</math> such that <math>a+b=c</math> is <math>\binom{6}{a+b} \sum_{a+b=c} \binom{6}{a}\binom{6}{b}</math> which by Vandermonde's is <math>\binom{6}{a+b}\binom{12}{a+b}</math>. Remember, that expression is the resulting sum for a fixed <math>a+b</math>. So, for <math>a+b\le 6</math>, we want <math>\sum_{c=0}^{6} \binom{6}{c}\binom{12}{c}</math>. This is (by Vandermonde's or committee forming) <math>\binom{18}{6} = 18564 \implies \boxed{564}</math> ~ firebolt360 | ||
+ | ===Note=== | ||
+ | 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 (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> | ||
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- Awsomness2000 | - Awsomness2000 | ||
− | ==Solution 3 (Major Bash)== | + | ==Solution 3 (Major Major Bash)== |
Case 1: <math>a<b</math>. | Case 1: <math>a<b</math>. | ||
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− | <cmath>316+124+124=\boxed{564}</cmath> | + | <cmath>316+124+124=\boxed{\boxed{564}}</cmath> |
==Solution 4== | ==Solution 4== | ||
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- mathymath | - mathymath | ||
− | ==Solution 5== | + | ==Solution 5 (Committee Forming but different)== |
− | Let | + | 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>. |
− | <math>6</math> | + | |
− | <math>6</math> | + | ==Solution 5 but different (Committee Forming)== |
− | <math>6</math> | + | Since <math>\binom{6}{n}=\binom{6}{6-n}</math>, we can rewrite <math>T(a,b)</math> as <math>\binom{6}{a}\binom{6}{b}\binom{6}{6-(a+b)}</math>. 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 <math>a</math> democrats, then pick <math>b</math> republicans, provided that <math>a+b \leq 6</math>. Then we can pick the remaining <math>6-(a+b)</math> people from the independents. But this is just <math>T(a,b)</math>, so the sum of all <math>T(a,b)</math> is equal to the number of ways to choose this committee. |
− | + | On the other hand, we can simply pick any 6 people from the <math>6+6+6=18</math> total politicians in the group. Clearly, there are <math>\binom{18}{6}</math> ways to do this. So the desired quantity is equal to <math>\binom{18}{6}</math>. We can then compute (routinely) the last 3 digits of <math>\binom{18}{6}</math> as <math>\boxed{564}</math>. | |
− | |||
− | |||
− | |||
− | |||
+ | ==Solution 6(NICE Journal) == | ||
+ | Note that <math>\binom{6}{a+b} = \binom{6}{6-a-b}</math>. So we have <math>\binom{6}{a}\binom{6}{b}\binom{6}{6-a-b}</math>. If we think about this this is essentially choosing a group of <math>a</math> people from <math>6</math> people, a group of <math>b</math> people from <math>6</math> people, and a group of <math>6 - a -b</math> from another group of <math>6</math> people. This is nothing but choosing <math>6</math> people from a group of <math>18</math> people. This is nothing but <math>\binom{18}{6} = 18564 \Rightarrow 564</math>. | ||
+ | ~coolmath_2018 | ||
+ | ==Remark == | ||
+ | This problem is an example of the generalization of Vandermonde's theorem, which states that for nonnegative <math>k_1, k_2, \ldots k_p</math> and <math>n_1, n_2, \ldots n_p</math>, we have | ||
+ | <cmath>\sum _{k_1+\cdots +k_p=m}\binom{n_1}{k_1} \binom{n_2}{k_2} \cdots \binom{n_p}{k_p} = \binom{n_1+ \cdots +n_p}{m}.</cmath> | ||
+ | ~eibc | ||
==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}} |
Latest revision as of 19:51, 29 January 2024
Contents
- 1 Problem 7
- 2 Major Note
- 3 Solution 1 (Committee Forming)
- 4 Solution 1 but different (Committee Forming)
- 5 Solution 2 (Committee Forming but slightly more bashy)
- 6 Solution 3 (Major Major Bash)
- 7 Solution 4
- 8 Solution 5 (Committee Forming but different)
- 9 Solution 5 but different (Committee Forming)
- 10 Solution 6(NICE Journal)
- 11 Remark
- 12 See Also
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 (Committee Forming but different)
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 5 but different (Committee Forming)
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
.
Solution 6(NICE Journal)
Note that . So we have
. If we think about this this is essentially choosing a group of
people from
people, a group of
people from
people, and a group of
from another group of
people. This is nothing but choosing
people from a group of
people. This is nothing but
.
~coolmath_2018
Remark
This problem is an example of the generalization of Vandermonde's theorem, which states that for nonnegative and
, we have
~eibc
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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