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

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- Apr 6th 2010, 12:15 PMranyengCoefficient 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 PMPlato
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 PMranyeng
Can you please explain with the above example( better if you write the steps)

- Apr 6th 2010, 01:05 PMPlato
- Apr 6th 2010, 01:43 PMawkward
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.)