Find the coefficient of $x^{20}$ in $(x + x^2 + x^3 + x^4 + x^5)(x^2 + x^3 + x^4 + \cdots)^5$ 
Find the coefficient of $x^{20}$ in $$(x + x^2 + x^3 + x^4 + x^5)(x^2 + x^3 + x^4 + \cdots)^5$$

I have transformed it into 
$$x^{11} (1+x+x^2+x^3+x^4) \left( \frac{1}{1-x} \right)^5$$
So I got an idea that from $x^{11}(1+x+x^2+x^3+x^4)$ we can get $x^{15},x^{14}, x^{13}, x^{12}, x^{11}$ and then on the right side we should have respectively $x^5, x^6, x^7, x^8 , x^9$. Am I right?
Having $$\left( \frac{1}{1-x} \right)^5 = \sum_{n=0}^\infty {5+n-1\choose 5-1}x^{n}$$
the final result that I got is
$$ {9\choose 4} + {10\choose 4} + {11\choose 4} + {12\choose 4} + {13\choose 4}$$
Is it correct?

Yes, it is correct. Another way where the final result is the difference of TWO binomial coefficients (rather than  the sum of FIVE ones): we have 
$$(x+x^2 +x^3 +x^4 +x^5)(x^2 +x^3 +x^4 +\dots)^5=\frac{x-x^6}{1-x}\cdot\frac{x^{10}}{(1-x)^5}=\frac{x^{11}}{(1-x)^6}-\frac{x^{16}}{(1-x)^6}$$
and it follows that
$$[x^{20}]\dots=[x^{9}]\frac{1}{(1-x)^6}-[x^{4}]\frac{1}{(1-x)^6}
=\binom{6-1+9}{6-1}-\binom{6-1+4}{6-1}=\binom{14}{5}-\binom{9}{5}.$$
