Determine the number of terms in simplified expression I've just started teaching myself algebra from a high school text-book and I stumbled upon this problem :

How many terms does the simplified form of $(a+b+c)^{2006}+(a-b-c)^{2006}$ have ? 

I found something on Wikipedia that appeared to be useful, the Multinomial Theorem, but It still led me to no result . Can somebody walk me through the solution ?
I'm much more interested in the thinking rather than the actual number.
 A: Hint. You need to calculate how many different terms of form $a^nb^mc^k$ can arise from expanding the power. Next you need to figure out what terms do cancel each other (one from $(a + b + c)^{2006}$ and the other from $(a - b- c)^{2006}$) and how many are those.
Edit. That hint was not very helpful. So, actually we need to count how many possible nonegative integer values $m, n, k$ satisfy
$$
m + n + k = 2006\\
(-1)^m(-1)^k = 1 \Leftrightarrow m + k \text{ is even}
$$
Since $m + k + n = 2006$, $m+k$ is even iff $n$ is even, i.e. $n = 2l$.
Finally, the problem is to count how many solutions does the following equation have
$$
2l + m + k = 2006, \qquad l,m,k \geq 0
$$
For a fixed $l$ that equation has $2007 - 2l$ solutions. So the total number of solutions will be
$$
\sum_{l=0}^{1003} 2007 - 2l = 1004 \cdot 2007 - 2 \frac{1003 \cdot 1004}{2} = 1004^2
$$
A: You don't need the multinomial theorem right away. What you need to notice is that $(a+b+c)=a + (b + c)$ and $(a-b-c)=a-(b+c)$. This means that when you do the simple binomial expansion, say $x = a$ and $y = b+c$, you lose a bunch of terms. To be precise, you lose all the terms that would have an odd power of $y$. So now you are left with expanding the remaining $1004$ terms (you can check this number). The remaining terms get doubled. 
So now you have $\sum_{i=0}^{1003} 2 {2006 \choose 2i} x^{2006-2i}y^{2i}$. I'm not quite sure how to complete it from here but I'm sure some algebraic manipulation (with some wrestling) would so the trick!
