, 2011

No three chips of the same colour should be in the same pile, so each pile must contain a chip of each colour. So essentially we need to partition $2n$ chips into two piles of $n$ each. This can be done in $p_n = \binom{2n}{n}/2$ ways.

Now,
$$\begin{align*} p_n &= \frac{1}{2} \binom{2n}{n} = \frac{(2n)!}{2(n!)^2} \\ &= 1 \cdot 3 \cdot \dotsb (2n-1) \frac{2 \cdot 4 \cdot \dotsb 2n}{2(n!)^2} \\ &= 1 \cdot 3 \cdot \dotsb (2n-1) \frac{2^n n!}{2(n!)^2} \\ &= 1 \cdot 3 \cdot \dotsb (2n-1) \frac{2^{n-1}}{n!}. \end{align*}$$

Now, let $m$ be a positive integer such that $2^m \le n < 2^{m+1}$. Then the greatest integer $a$ such that $2^a$ divides $n!$ is
$$\begin{align*} a &= \lfloor \frac{n}{2} \rfloor + \lfloor \frac{n}{2^2} \rfloor + \lfloor \frac{n}{2^3} \rfloor + \dots + \lfloor \frac{n}{2^m} \rfloor \\ &\le \frac{n}{2} + \frac{n}{2^2} + \frac{n}{2^3} + \dots + \frac{n}{2^m} \\ &= n - \frac{n}{2^m} \le n-1. \end{align*}$$

Now, $p_n$ is odd exactly when $a = n-1$, which happens if only if $n = 2^m$.