Problem: The expression $ (x+y+z)^{2006}+(x-y-z)^{2006} $ is simplified by expanding it and combining like terms. How many terms are in the simplified expression? Solution: First, notice by balls and bins that there are $\binom{2006 + 3 - 1}{3 - 1} = \binom{2008}{2}$ terms in $(x+y+z)^{2006}$. We want to know how many terms are *not* canceled by $(x-y-z)^{2006}$. These are simply the terms that are added instead of subtracted. So, we ask, when is a term in $(x-y-z)^{2006}$ going to have a negative coefficient? Say, the number $r_{a,b,c}$ denotes the coefficient of the term in the expansion of $(a-y-z)^{2006}$ of the $x^ay^bz^c$ term. Notice that $r_{a,b,c}$ is positive when $b+c = 2k$ for some $k \in \mathbb{N}$. We see that $a = 2a_{0}$ for some $m \in \mathbb{N}$ in this case and so if $b+c=2k$ then either $b,c$ are both even or both odd. Assume, $b,c$ are both even, then, we have that $b = 2b_{0}$ and $c = 2c_{0}$ for some $b_{0},c_{0} \in \mathbb{N}^2$ and so we want to find all integral solutions to $a_{0}+b_{0}+c_{0} = 1003$ which is simply $\binom{1005}{2}$. If $b,c$ are both odd, then $b = 2b'+1$ and $c = 2c' + 1$ so $a+b+c = 2a_{0}+2b'+2c'+2 = 2006$ so it suffices to find all integral solutions to $a_{0}+b'+c' = 1002$ which is just $\binom{1004}{2}$. Thus, there are $\binom{1005}{2}+\binom{1004}{2} = 1008016$ total terms in the expression.