Finding Coefficient of x^8y^5 using Binomial Theorem

[reenmachine]

Homework Statement

Use the binomial theorem to find the coefficient of $x^8y^5$ in $(x+y)^{13}$.

Homework Equations

We know 13 - 5 = 8 , so we have $\binom{n}{5}x^{n-5}y^5 = \binom{13}{5}x^8y^5$

$\binom{13}{5} = \frac{13 \cdot 12 \cdot 11 \cdot 10 \cdot 9 \cdot 8!}{5!8!} = \frac{13 \cdot 12 \cdot 11 \cdot 10 \cdot 9}{120} = \frac{154440}{120} = 1287$

So $1287x^8y^5$

This is the first time I work with the binomial theorem so I'm not sure , any thoughts on my result?

thank you!

 

[mfb]

I don't see why you introduced "n" there.
The solution is right.

 

[reenmachine]

I don't see why you introduced "n" there.
The solution is right.

sorry this was a brain cramp on my part.The book I'm reading introduced the binomial theorem as followed: If $n$ is a non-negative integer , then $(x+y)^n = \binom{n}{0}x^n + \binom{n}{1}x^{n-1}y + \binom{n}{2}x^{n-2}y^2 + \binom{n}{3}y^3 + \cdots + \binom{n}{n-1}xy^{n-1}+\binom{n}{n}y^n$.For some reasons I forgot to connect $n$ to $(x+y)^n$ and $\mathbb{N}$.

thank you!