2006 AMC 12: how many terms in the simplified expression?

[malick]

Hello, this is a question from the 2006 AMC 12:

The expression "(x + y + z)²⁰⁰⁶ + (z - y - z)²⁰⁰⁶" is simplified by expanding it and combining like terms. How many terms are in the simplified expression?

 

[stapel]

What rules do you know about expanding polynomials of this type?

What can you say about the number of terms in (x + y + z)²⁰⁰⁶ and (z - y - x)²⁰⁰⁶ which will be the same but for their sign (and thus cancel out when you combine like terms)?

Please reply showing what you have tried thus far, and what your thoughts are.

Thank you.

Eliz.

 

[malick]

I know that there are 2006+1 (2007) values to (x+y+z)^2006 Each of these 2007 values corresponds to exactly one term in the expansion. This is about it

 

[pka]

I know that there are 2006+1 (2007) values to (x+y+z)^2006 Each of these 2007 values corresponds to exactly one term in the expansion. This is about it

Oh my goodness, NO!
In \(\displaystyle (x+y+z)^{2006}\) there are 2,015,028 terms.

Each term looks like \(\displaystyle \frac{{2006!}}{{\left( {j!} \right)\left( {k!} \right)\left( {n!} \right)}}x^j y^k z^n ,\quad j + k + n = 2006\).

The number 2,015,028 comes from \(\displaystyle 2008 \choose 2\), that is the number of non-negative integer solutions to the equation \(\displaystyle j + k + n = 2006\) there are.

In \(\displaystyle (x-y-z)^{2006}\) there are 2,015,028 terms each term looks like
\(\displaystyle \frac{{2006!}}{{\left( {j!} \right)\left( {k!} \right)\left( {n!} \right)}}x^j (-y)^k (-z)^n ,\quad j + k + n = 2006\).

Thus terms with opposite signs will subtract off.

 

[galactus]

http://wangsblog.com/jeffrey/?cat=8