2006 AMC 12A Problems/Problem 24
Problem
The expression
is simplified by expanding it and combining like terms. How many terms are in the simplified expression?
Solution
by the multi-nomial theorem, the expressions of the two are:
$\sum{\frac{2006!}{a!b!c!}x^ay^bz^c}$
and:
$\sum{\frac{2006!}{a!b!c!}x^a(-y)^b(-z)^c}$
respectively, where $a+b+c = 2006$. 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:
$\binom{2006+2}{2} = 2015028$
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 1003 solutions.
if the exponent of $y$ is 3, the exponent of $z$ can go up to $2002$, so 1002 solutions.
$\vdots$
if the exponent of $y$ is 2005$, then $z$ can only be $0$. So 1 solution.
add them up we get $\frac{1003*1004}{2}$ solutions. 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 $1003*1004$ negative terms.
Subtract this number from 2015028 we obtain $D. 1008016$ as our answer.
