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Difference between revisions of "2015 AIME I Problems/Problem 12"
(→‎Solution)
Line 12: Line 12:
 
==Solution==
 
==Solution==
 
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,
 
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*}
 
\begin{align*}
 
\binom{2015}{1000} M &= 1 \cdot \binom{2014}{999} + 2 \cdot \binom{2013}{999} + \dots + 1016 \cdot \binom{999}{999} \\
 
\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{2014}{999} + \binom{2013}{999} + \dots + \binom{999}{999} \\
 
     &                      + \binom{2013}{999} + \binom{2012}{999} + \dots + \binom{999}{999} \\
 
     &                      + \binom{2013}{999} + \binom{2012}{999} + \dots + \binom{999}{999} \\
 
     & \dots \\
 
     & \dots \\
Line 21: Line 22:
 
     &= \binom{2016}{1001}.
 
     &= \binom{2016}{1001}.
 
\end{align*}
 
\end{align*}
 +
</cmath>
 
Using the definition of binomial coefficient and the identity <math>n! = n \cdot (n-1)!</math>, we deduce that
 
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>
 
<cmath>M = \frac{2016}{1001} = \frac{288}{143}.</cmath>

Revision as of 17:46, 20 March 2015

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

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 $\boxed{431}.$

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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