# Coefficient in the expansion

• Apr 6th 2010, 12:15 PM
ranyeng
Coefficient in the expansion
What is the coefficient of x^6 in the expansion (1+x+x^2)^9. Explain please
• Apr 6th 2010, 12:46 PM
Plato
Quote:

Originally Posted by ranyeng
What is the coefficient of x^6 in the expansion (1+x+x^2)^9. Explain please

Count the number of ways to rearrange each of these string.
$\displaystyle 111xxxxxx,~1111xxxxx^2,~ 11111xxx^2x^2,~ \&~111111x^2x^2x^2$
For example: the string $\displaystyle 1111xxxxx^2$ can be rearranged in $\displaystyle \frac{9!}{(1!)(4!)(4!)}$ ways.
Note that product is $\displaystyle x^6$.
The sum of the numbers of rearrangements is the coefficient of $\displaystyle x^6$.
• Apr 6th 2010, 01:00 PM
ranyeng
Can you please explain with the above example( better if you write the steps)
• Apr 6th 2010, 01:05 PM
Plato
Quote:

Originally Posted by ranyeng
Can you please explain with the above example( better if you write the steps)

No sorry that happens to be your job.

I will say the number of ways to rearrange the string $\displaystyle aabbbcccc$ is $\displaystyle \frac{9!}{(2!)(3!)(4!)}$.
If you do not understand that rule, then you have no business trying this question.
• Apr 6th 2010, 01:43 PM
awkward
Quote:

Originally Posted by ranyeng
What is the coefficient of x^6 in the expansion (1+x+x^2)^9. Explain please

Another approach is to make use of the identity

$\displaystyle (1+x+x^2)^9 = \left( \frac{1-x^3}{1-x} \right) ^9 = (1-x^3)^9 \; (1-x)^{-9}$

then expand $\displaystyle (1-x^3)^9$ and $\displaystyle (1-x)^{-9}$ by the binomial theorem.

(Assuming you know how to use the binomial theorem for negative exponents.)