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Difference between revisions of "2015 AIME I Problems/Problem 12"

(Created page with "==Problem== Consider all 1000-element subsets of the set {1, 2, 3, ... , 2015}. From each such subset choose the least element. The arithmetic mean of all of these least el...")
 
(→‎Solution 4 (short))
 
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==Problem==
 
==Problem==
  
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Consider all 1000-element subsets of the set {1, 2, 3, ... , 2015}.  From each such subset choose the least element.  The arithmetic mean of all of these least elements is <math> \frac{p}{q} </math>, where <math>p</math> and <math>q</math> are relatively prime positive integers.  Find <math>p + q</math>.
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Consider all 1000-element subsets of the set <math>\{1, 2, 3, ... , 2015\}</math>.  From each such subset choose the least element.  The arithmetic mean of all of these least elements is <math> \frac{p}{q} </math>, where <math>p</math> and <math>q</math> are relatively prime positive integers.  Find <math>p + q</math>.
 +
 
 +
==Hint==
 +
Use the Hockey Stick Identity in the form
 +
 
 +
<cmath>\binom{a}{a} + \binom{a+1}{a} + \binom{a+2}{a} + \dots + \binom{b}{a} = \binom{b+1}{a+1}.</cmath>
 +
 
 +
(This is best proven by a combinatorial argument that coincidentally pertains to the problem: count two ways the number of subsets of the first <math>(b + 1)</math> numbers with <math>(a + 1)</math> elements whose least element is <math>i</math>, for <math>1 \le i \le b - a</math>.)
 +
 
 +
==Solution==
 +
===Solution 1===
 +
Let <math>M</math> be the desired mean. Then because <math>\dbinom{2015}{1000}</math> subsets have 1000 elements and <math>\dbinom{2015 - i}{999}</math> have <math>i</math> as their least element,
 +
<cmath>
 +
\begin{align*}
 +
\binom{2015}{1000} M &= 1 \cdot \binom{2014}{999} + 2 \cdot \binom{2013}{999} + \dots + 1016 \cdot \binom{999}{999} \\
 +
    &= \binom{2014}{999} + \binom{2013}{999} + \dots + \binom{999}{999} \\
 +
    &                      + \binom{2013}{999} + \binom{2012}{999} + \dots + \binom{999}{999} \\
 +
    & \dots \\
 +
    & + \binom{999}{999} \\
 +
    &= \binom{2015}{1000} + \binom{2014}{1000} + \dots + \binom{1000}{1000} \\
 +
    &= \binom{2016}{1001}.
 +
\end{align*}
 +
</cmath>
 +
Using the definition of binomial coefficient and the identity <math>n! = n \cdot (n-1)!</math>, we deduce that
 +
<cmath>M = \frac{2016}{1001} = \frac{288}{143}.</cmath>
 +
The answer is <math>\boxed{431}.</math>
 +
 
 +
===Solution 2===
 +
Each 1000-element subset <math>\left\{ a_1, a_2,a_3,...,a_{1000}\right\}</math> of <math>\left\{1,2,3,...,2015\right\}</math> with <math>a_1<a_2<a_3<...<a_{1000}</math> contributes <math>a_1</math> to the sum of the least element of each subset. Now, consider the set <math>\left\{a_1+1,a_2+1,a_3+1,...,a_{1000}+1\right\}</math>. There are <math>a_1</math> ways to choose a positive integer <math>k</math> such that <math>k<a_1+1<a_2+1,a_3+1<...<a_{1000}+1</math> (<math>k</math> can be anything from <math>1</math> to <math>a_1</math> inclusive). Thus, the number of ways to choose the set <math>\left\{k,a_1+1,a_2+1,a_3+1,...,a_{1000}+1\right\}</math> is equal to the sum. But choosing a set <math>\left\{k,a_1+1,a_2+1,a_3+1,...,a_{1000}+1\right\}</math> is the same as choosing a 1001-element subset from <math>\left\{1,2,3,...,2016\right\}</math>!
 +
 
 +
Thus, the average is <math>\frac{\binom{2016}{1001}}{\binom{2015}{1000}}=\frac{2016}{1001}=\frac{288}{143}</math>. Our answer is <math>p+q=288+143=\boxed{431}</math>.
 +
 
 +
===Solution 3(NOT RIGOROUS)===
 +
Let <math>p</math> be the size of the large set and <math>q</math> be the size of the subset (i.e. in this problem, <math>p = 2015</math> and <math>q = 1000</math>). We can easily find the answers for smaller values of <math>p</math> and <math>q</math>:
 +
 
 +
For <math>p = 2</math> and <math>q = 2</math>, the answer is <math>1</math>.
 +
 
 +
For <math>p = 3</math> and <math>q = 2</math>, the answer is <math>\frac43</math>.
 +
 
 +
For <math>p = 4</math> and <math>q = 2</math>, the answer is <math>\frac53</math>.
 +
 
 +
For <math>p = 3</math> and <math>q = 3</math>, the answer is <math>1</math>.
 +
 
 +
For <math>p = 4</math> and <math>q = 3</math>, the answer is <math>\frac54</math>.
 +
 
 +
For <math>p = 5</math> and <math>q = 3</math>, the answer is <math>\frac32</math>.
 +
 
 +
At this point, we can see a pattern: our desired answer is always <math>\frac{p+1}{q+1}</math>. Plugging in <math>p = 2015</math> and <math>q = 1000</math>, the answer is <math>\frac{2016}{1001}=\frac{288}{143}</math>, so <math>288 + 143 = \boxed{431}</math>.
 +
 
 +
===Solution 4 (short)===
 +
In the "average case", the numbers evenly partition the interval [0,2016] into 1001 parts. Then because it asks for the expected value of the least element the answer is <math>\frac{2016}{1001}</math>.
 +
 
 +
-tigershark22
 +
 
 +
VIEWER NOTE: This solution doesn't always work, for example, take <math>n^2</math> on <math>[0,1]</math>. The "average case" is <math>n=\frac{1}{2}\implies n^2=\frac{1}{4}</math> but integrating <math>n^2</math> from <math>0</math> to <math>1</math> we see that definitely is not the case. This works only in certain situations, so maybe this solution would be better off with proof that this is one of those situations.
 +
 
 +
==Video Solution==
 +
 
 +
https://www.youtube.com/watch?v=t-yHTkGkMK4
 +
~anellipticcurveoverq
 +
 
 +
== Generalization ==
 +
https://artofproblemsolving.com/wiki/index.php/1981_IMO_Problems/Problem_2
 +
 
 +
== See also ==
 +
{{AIME box|year=2015|n=I|num-b=11|num-a=13}}
 +
{{MAA Notice}}
 +
[[Category:Introductory Combinatorics Problems]]

Latest revision as of 22:01, 9 August 2021

Problem

Consider all 1000-element subsets of the set $\{1, 2, 3, ... , 2015\}$. From each such subset choose the least element. The arithmetic mean of all of these least elements is $\frac{p}{q}$, where $p$ and $q$ are relatively prime positive integers. Find $p + q$.

Hint

Use the Hockey Stick Identity in the form

\[\binom{a}{a} + \binom{a+1}{a} + \binom{a+2}{a} + \dots + \binom{b}{a} = \binom{b+1}{a+1}.\]

(This is best proven by a combinatorial argument that coincidentally pertains to the problem: count two ways the number of subsets of the first $(b + 1)$ numbers with $(a + 1)$ elements whose least element is $i$, for $1 \le i \le b - a$.)

Solution

Solution 1

Let $M$ be the desired mean. Then because $\dbinom{2015}{1000}$ subsets have 1000 elements and $\dbinom{2015 - i}{999}$ have $i$ as their least element, \begin{align*} \binom{2015}{1000} M &= 1 \cdot \binom{2014}{999} + 2 \cdot \binom{2013}{999} + \dots + 1016 \cdot \binom{999}{999} \\ &= \binom{2014}{999} + \binom{2013}{999} + \dots + \binom{999}{999} \\ & + \binom{2013}{999} + \binom{2012}{999} + \dots + \binom{999}{999} \\ & \dots \\ & + \binom{999}{999} \\ &= \binom{2015}{1000} + \binom{2014}{1000} + \dots + \binom{1000}{1000} \\ &= \binom{2016}{1001}. \end{align*} Using the definition of binomial coefficient and the identity $n! = n \cdot (n-1)!$, we deduce that \[M = \frac{2016}{1001} = \frac{288}{143}.\] The answer is $\boxed{431}.$

Solution 2

Each 1000-element subset $\left\{ a_1, a_2,a_3,...,a_{1000}\right\}$ of $\left\{1,2,3,...,2015\right\}$ with $a_1<a_2<a_3<...<a_{1000}$ contributes $a_1$ to the sum of the least element of each subset. Now, consider the set $\left\{a_1+1,a_2+1,a_3+1,...,a_{1000}+1\right\}$. There are $a_1$ ways to choose a positive integer $k$ such that $k<a_1+1<a_2+1,a_3+1<...<a_{1000}+1$ ($k$ can be anything from $1$ to $a_1$ inclusive). Thus, the number of ways to choose the set $\left\{k,a_1+1,a_2+1,a_3+1,...,a_{1000}+1\right\}$ is equal to the sum. But choosing a set $\left\{k,a_1+1,a_2+1,a_3+1,...,a_{1000}+1\right\}$ is the same as choosing a 1001-element subset from $\left\{1,2,3,...,2016\right\}$!

Thus, the average is $\frac{\binom{2016}{1001}}{\binom{2015}{1000}}=\frac{2016}{1001}=\frac{288}{143}$. Our answer is $p+q=288+143=\boxed{431}$.

Solution 3(NOT RIGOROUS)

Let $p$ be the size of the large set and $q$ be the size of the subset (i.e. in this problem, $p = 2015$ and $q = 1000$). We can easily find the answers for smaller values of $p$ and $q$:

For $p = 2$ and $q = 2$, the answer is $1$.

For $p = 3$ and $q = 2$, the answer is $\frac43$.

For $p = 4$ and $q = 2$, the answer is $\frac53$.

For $p = 3$ and $q = 3$, the answer is $1$.

For $p = 4$ and $q = 3$, the answer is $\frac54$.

For $p = 5$ and $q = 3$, the answer is $\frac32$.

At this point, we can see a pattern: our desired answer is always $\frac{p+1}{q+1}$. Plugging in $p = 2015$ and $q = 1000$, the answer is $\frac{2016}{1001}=\frac{288}{143}$, so $288 + 143 = \boxed{431}$.

Solution 4 (short)

In the "average case", the numbers evenly partition the interval [0,2016] into 1001 parts. Then because it asks for the expected value of the least element the answer is $\frac{2016}{1001}$.

-tigershark22

VIEWER NOTE: This solution doesn't always work, for example, take $n^2$ on $[0,1]$. The "average case" is $n=\frac{1}{2}\implies n^2=\frac{1}{4}$ but integrating $n^2$ from $0$ to $1$ we see that definitely is not the case. This works only in certain situations, so maybe this solution would be better off with proof that this is one of those situations.

Video Solution

https://www.youtube.com/watch?v=t-yHTkGkMK4 ~anellipticcurveoverq

Generalization

https://artofproblemsolving.com/wiki/index.php/1981_IMO_Problems/Problem_2

See also

2015 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 11 		Followed by
Problem 13
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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