Find the coefficient of $x^{20}$ in $(x^{1}+⋯+x^{6} )^{10}$ I'm trying to find the coefficient of $x^{20}$ in 
$$(x^{1}+⋯+x^{6} )^{10}$$
So I did this : 
$$\frac {1-x^{m+1}} {1-x} = 1+x+x^2+⋯+x^{m}$$
$$(x^1+⋯+x^6 )=x(1+x+⋯+x^5 ) = \frac {x(1-x^6 )} {1-x} = \frac {x-x^7} {1-x}$$
$$(x^1+⋯+x^6 )^{10} =\left(\dfrac {x-x^7} {1-x}\right)^{10}$$
But what do I do from here ? any hints ? 
Thanks 

You can factor out $x^{10}$ so that your goal becomes to find the $x^{10}$ coefficient in $(1 + x + .. + x^5)^{10}$. By the Multinomial Theorem  one has
$$(1 + x + .. + x^5)^{10} = \sum_{k_1 + ... + k_6 = 10} \frac{10!}{k_1!k_2!k_3!k_4! k_5!k_6} x^{k_2 + 2k_3 + 3k_4 + 4k_5 + 5k_6}$$
Here the sum is over nonnegative integers. So what you need to do is find all $(k_1,...,k_6)$ such that $k_2 + 2k_3 + 3k_4 + 4k_5 + 5k_6 = 10$, or equivalently all 
$(k_2,...,k_6)$ such that such that $k_2 + 2k_3 + 3k_4 + 4k_5 + 5k_6 \leq 10$, then add up the corresponding coefficients.
