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Difference between revisions of "2006 Canadian MO Problems/Problem 1"
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Let <math>f(n,k)</math> be the number of ways distributing <math>k</math> candies to <math>n</math> children so that each child receives at most two candies. For example, <math>f(3,7)=0</math>, <math>f(3,6)=1</math>, and <math>f(3,4)=6</math>. Evaluate <math>f(2006,1)+f(2006,4)+f(2006,7)+\dots+f(2006,1003)</math>.
 
Let <math>f(n,k)</math> be the number of ways distributing <math>k</math> candies to <math>n</math> children so that each child receives at most two candies. For example, <math>f(3,7)=0</math>, <math>f(3,6)=1</math>, and <math>f(3,4)=6</math>. Evaluate <math>f(2006,1)+f(2006,4)+f(2006,7)+\dots+f(2006,1003)</math>.
 
==Solution==
 
==Solution==
−
\[
 
−
f(n, k) = \text{the number of ways to distribute } k \text{ candies among } n \text{ children such that each child gets at most 2 candies. This corresponds to finding non-negative integer solutions to } x_1 + x_2 + \cdots + x_n = k \text{ where } 0 \leq x_i \leq 2 \text{ for all } i. \text{ The generating function for one child is } 1 + x + x^2, \text{ and for } n \text{ children, it becomes } (1 + x + x^2)^n. \text{ Rewriting, } 1 + x + x^2 = \frac{1 - x^3}{1 - x}, \text{ so } (1 + x + x^2)^n = \left(\frac{1 - x^3}{1 - x}\right)^n = (1 - x^3)^n (1 - x)^{-n}. \text{ Using the binomial theorem, we expand } (1 - x^3)^n = \sum_{j=0}^n \binom{n}{j} (-1)^j x^{3j} \text{ and } (1 - x)^{-n} = \sum_{m=0}^\infty \binom{n + m - 1}{m} x^m. \text{ The coefficient of } x^k \text{ in } (1 + x + x^2)^n \text{ is given by } f(n, k) = \sum_{j=0}^{\lfloor k/3 \rfloor} \binom{n}{j} (-1)^j \binom{n + k - 3j - 1}{k - 3j}. \text{ For the problem, we sum } f(2006, k) \text{ over odd } k \text{ from } 1 \text{ to } 1003. \text{ By symmetry of the generating function, } \sum_{k \text{ odd}} f(n, k) = 2^{n-1}. \text{ Thus, for } n = 2006, \text{ the sum becomes } \sum_{k \text{ odd}} f(2006, k) = 2^{2006 - 1} = 2^{2005}. \text{ Therefore, the final answer is } \boxed{2^{2005}}.
 
−
\]
 
  
 +
<math>f(n, k) = \text{the number of ways to distribute } k \text{ candies among } n \text{ children such that each child gets at most 2 candies.} </math>
 +
 +
<math> \text{This corresponds to finding non-negative integer solutions to } x_1 + x_2 + \cdots + x_n = k \text{ where } 0 \leq x_i \leq 2 \text{ for all } i.</math>
 +
 +
<math>\text{ The generating function for one child is } 1 + x + x^2, \text{ and for } n \text{ children, it becomes } (1 + x + x^2)^n. </math>
 +
 +
<math>\text{ Rewriting, } 1 + x + x^2 = \frac{1 - x^3}{1 - x}, \text{ so } (1 + x + x^2)^n = \left(\frac{1 - x^3}{1 - x}\right)^n = (1 - x^3)^n (1 - x)^{-n}. </math>
 +
 +
<math> \text{ Using the binomial theorem, we expand } (1 - x^3)^n = \sum_{j=0}^n \binom{n}{j} (-1)^j x^{3j} \text{ and } (1 - x)^{-n} = \sum_{m=0}^\infty \binom{n + m - 1}{m} x^m. </math>
 +
 +
<math>\text{ The coefficient of } x^k \text{ in } (1 + x + x^2)^n \text{ is given by } f(n, k) = \sum_{j=0}^{\lfloor k/3 \rfloor} \binom{n}{j} (-1)^j \binom{n + k - 3j - 1}{k - 3j}. </math>
 +
 +
<math> \text{ For the problem, we sum } f(2006, k) \text{ over odd } k \text{ from } 1 \text{ to } 1003. </math>
 +
 +
<math>\text{ By symmetry of the generating function, } \sum_{k \text{ odd}} f(n, k) = 2^{n-1}. </math>
 +
 +
<math>\text{ Thus, for } n = 2006, \text{ the sum becomes } \sum_{k \text{ odd}} f(2006, k) = 2^{2006 - 1} = 2^{2005}. </math>
 +
 +
<math>\text{ Therefore, the final answer is } \boxed{2^{2005}}. </math>
 +
 +
~sitar
  
 
==See also==
 
==See also==

Latest revision as of 18:24, 17 November 2024

Problem

Let $f(n,k)$ be the number of ways distributing $k$ candies to $n$ children so that each child receives at most two candies. For example, $f(3,7)=0$, $f(3,6)=1$, and $f(3,4)=6$. Evaluate $f(2006,1)+f(2006,4)+f(2006,7)+\dots+f(2006,1003)$.

Solution

$f(n, k) = \text{the number of ways to distribute } k \text{ candies among } n \text{ children such that each child gets at most 2 candies.}$

$\text{This corresponds to finding non-negative integer solutions to } x_1 + x_2 + \cdots + x_n = k \text{ where } 0 \leq x_i \leq 2 \text{ for all } i.$

$\text{ The generating function for one child is } 1 + x + x^2, \text{ and for } n \text{ children, it becomes } (1 + x + x^2)^n.$

$\text{ Rewriting, } 1 + x + x^2 = \frac{1 - x^3}{1 - x}, \text{ so } (1 + x + x^2)^n = \left(\frac{1 - x^3}{1 - x}\right)^n = (1 - x^3)^n (1 - x)^{-n}.$

$\text{ Using the binomial theorem, we expand } (1 - x^3)^n = \sum_{j=0}^n \binom{n}{j} (-1)^j x^{3j} \text{ and } (1 - x)^{-n} = \sum_{m=0}^\infty \binom{n + m - 1}{m} x^m.$

$\text{ The coefficient of } x^k \text{ in } (1 + x + x^2)^n \text{ is given by } f(n, k) = \sum_{j=0}^{\lfloor k/3 \rfloor} \binom{n}{j} (-1)^j \binom{n + k - 3j - 1}{k - 3j}.$

$\text{ For the problem, we sum } f(2006, k) \text{ over odd } k \text{ from } 1 \text{ to } 1003.$

$\text{ By symmetry of the generating function, } \sum_{k \text{ odd}} f(n, k) = 2^{n-1}.$

$\text{ Thus, for } n = 2006, \text{ the sum becomes } \sum_{k \text{ odd}} f(2006, k) = 2^{2006 - 1} = 2^{2005}.$

$\text{ Therefore, the final answer is } \boxed{2^{2005}}.$

~sitar

See also

2006 Canadian MO (Problems)
Preceded by
First question 		1 • 2 • 3 • 4 • 5 Followed by
Problem 2
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