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.

See also

• 2006 AMC 12A Problems