2006 Canadian MO Problems/Problem 1
Problem
Let 
be the number of ways distributing 
candies to 
children so that each child receives at most two candies. For example, 
, 
, and 
. Evaluate 
.
Solution
\[ f(n, k) = \text{the number of ways to distribute } k \text{ candies among } n \text{ children such that each child gets at most 2 candies. This corresponds to finding non-negative integer solutions to } x_1 + x_2 + \cdots + x_n = k \text{ where } 0 \leq x_i \leq 2 \text{ for all } i. \text{ The generating function for one child is } 1 + x + x^2, \text{ and for } n \text{ children, it becomes } (1 + x + x^2)^n. \text{ Rewriting, } 1 + x + x^2 = \frac{1 - x^3}{1 - x}, \text{ so } (1 + x + x^2)^n = \left(\frac{1 - x^3}{1 - x}\right)^n = (1 - x^3)^n (1 - x)^{-n}. \text{ Using the binomial theorem, we expand } (1 - x^3)^n = \sum_{j=0}^n \binom{n}{j} (-1)^j x^{3j} \text{ and } (1 - x)^{-n} = \sum_{m=0}^\infty \binom{n + m - 1}{m} x^m. \text{ The coefficient of } x^k \text{ in } (1 + x + x^2)^n \text{ is given by } f(n, k) = \sum_{j=0}^{\lfloor k/3 \rfloor} \binom{n}{j} (-1)^j \binom{n + k - 3j - 1}{k - 3j}. \text{ For the problem, we sum } f(2006, k) \text{ over odd } k \text{ from } 1 \text{ to } 1003. \text{ By symmetry of the generating function, } \sum_{k \text{ odd}} f(n, k) = 2^{n-1}. \text{ Thus, for } n = 2006, \text{ the sum becomes } \sum_{k \text{ odd}} f(2006, k) = 2^{2006 - 1} = 2^{2005}. \text{ Therefore, the final answer is } \boxed{2^{2005}}. \]
See also
| 2006 Canadian MO (Problems) | ||
| Preceded by First question |
1 • 2 • 3 • 4 • 5 | Followed by Problem 2 |
