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== | ||
We begin as in solution 1 to rewrite the sum as <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>. | We begin as in solution 1 to rewrite the sum as <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 the polynomial <math>P(x)=(\binom{6}{0} + \binom{6}{1}x + \binom{6}{2}x^2 + \binom{6}{3}x^3 + \cdot \cdot \cdot + \binom{6}{6}x^6)^3</math>. | + | Consider the polynomial <math>P(x)=\left(\binom{6}{0} + \binom{6}{1}x + \binom{6}{2}x^2 + \binom{6}{3}x^3 + \cdot \cdot \cdot + \binom{6}{6}x^6\right)^3</math>. |
We can see the sum we wish to compute is just the coefficient of the <math>x^6</math> term. However <math>P(x)=((1+x)^6)^3=(1+x)^{18}</math>. Therefore, the coefficient of the <math>x^6</math> term is just <math>\binom{18}{6} = 18564</math> so the answer is <math>\boxed{564}</math>. | We can see the sum we wish to compute is just the coefficient of the <math>x^6</math> term. However <math>P(x)=((1+x)^6)^3=(1+x)^{18}</math>. Therefore, the coefficient of the <math>x^6</math> term is just <math>\binom{18}{6} = 18564</math> so the answer is <math>\boxed{564}</math>. | ||
- mathymath | - mathymath | ||
+ | |||
+ | ==Solution 5 (Committee Forming but different)== | ||
+ | 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>. | ||
+ | |||
+ | ==Solution 5 but different (Committee Forming)== | ||
+ | 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 19:01, 26 February 2023
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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