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?
$\mathrm{(A) \ } 6018\qquad \mathrm{(B) \ } 671,676\qquad \mathrm{(C) \ } 1,007,514\qquad \mathrm{(D) \ } 1,008,016\qquad\mathrm{(E) \ }  2,015,028$
By the [Multinomial Theorem](https://artofproblemsolving.com/wiki/index.php/Multinomial_Theorem), the summands can be written as
\[\sum_{a+b+c=2006}{\frac{2006!}{a!b!c!}x^ay^bz^c}\]
and
\[\sum_{a+b+c=2006}{\frac{2006!}{a!b!c!}x^a(-y)^b(-z)^c},\]
respectively. Since the coefficients of like terms are the same in each expression, each like term either cancel one another out or the coefficient doubles. In each expansion there are:
\[{2006+2\choose 2} = 2015028\]
terms without cancellation. For any term in the second expansion to be negative, the parity of the exponents of $y$ and $z$ must be opposite. Now we find a pattern:
if the exponent of $y$ is $1$, the exponent of $z$ can be all even integers up to $2004$, so there are $1003$ terms.
if the exponent of $y$ is $3$, the exponent of $z$ can go up to $2002$, so there are $1002$ terms.
$\vdots$
if the exponent of $y$ is $2005$, then $z$ can only be 0, so there is $1$ term.
If we add them up, we get $\frac{1003\cdot1004}{2}$ terms. However, we can switch the exponents of $y$ and $z$ and these terms will still have a negative sign. So there are a total of $1003\cdot1004$ negative terms.
By subtracting this number from 2015028, we obtain $\boxed{\mathrm{D}}$ or $1,008,016$ as our answer.