How to determine the coefficient of $x^{10}$ in the expansion $(1+x+x^2+x^3+.....+x^{10})^4$ I have a question

Find the coefficient of $x^{10}$ in the expansion $(1+x+x^2+x^3+.....+x^{10})^4$

There ARE questions like this on stack exchange already I know, but I'm not able to formulate a pattern or know how to apply that thing here... I've tried making combinations of $1$'s and $x^{10}$'s, $x$'s and $x^9$'s etc but I am unable to solve it. Please help.
PS. How to do it using combinations exclusively.
 A: Try:
$\begin{align*}
1 + x + &x^2 + x^3 + x^4 + x^5 + x^6 + x^7 + x^8 + x^9 + x^{10} \\
  &= \frac{1 - x^{11}}{1- x} \\
(1 + x + &x^2 + x^3 + x^4 + x^5 + x^6 + x^7 + x^8 + x^9 + x^{10})^4 \\
  &= \frac{(1 - x^{11})^4}{(1- x)^4} \\
  &= (1 - 4 x^{11} + 6 x^{22} - 4 x^{33} + x^{44})
          \cdot \sum_{k \ge 0} (-1)^k \binom{-4}{k} x^k \\
[x^{10}]  (1 - &4 x^{11} + 6 x^{22} - 4 x^{33} + x^{44})
          \cdot \sum_{k \ge 0} (-1)^k \binom{-4}{k} x^k \\
  &= [x^{10}] \sum_{k \ge 0} (-1)^k \binom{-4}{k} x^k \\
  &= (-1)^{10} \binom{-4}{10} \\
  &= \binom{10 + 4 - 1}{4 - 1} \\
  &= 286
\end{align*}$
A: The standard geometric series tells us that
$$1 +x + x^2 + x^3 + \ldots +x^{10}= \frac{1-x^{11}}{1-x}$$
So taking this to the fourth power we get 
$$\left( \frac{1-x^{11}}{1-x} \right)^4 = (1-x^{11})^4 (1-x)^{-4}\tag{1}$$
This would not seem to really help much, except that for negative powers of $n$ there
is a generalised binomial formula:
$$(1-x)^{-4} = \sum_{k=0}^\infty \binom{k+3}{k}x^k$$
The standard binomial formula gives us for the positive power $4$ that:
$$(1-x^{11})^{4} = \sum_{k=0}^{4} \binom{4}{k}(-1)^k x^{11k}$$
So $(1)$ becomes
$$\left(1 -\binom{4}{1}x^{11} + \binom{4}{2}x^{22} - \binom{4}{3}x^{33} + \binom{4}{4}x^{44} \right)\left(\binom{3}{0} +\binom{4}{1}x + \binom{5}{2}x^2 + \ldots\right)\tag{2}$$
And we still want the coefficient of $x^{10}$. But terms with $x^{11}$ or more we cannot use from the left hand side when multiplying out, so we only get the term where we use $1$ from the left hand sum and $\binom{13}{10}$ (the coefficient of $x^{10}$) from the right hand sum. 
So the answer is $\binom{13}{10} = 286$. Thanks to the (generalised) binomial formula and geometric series.
A: When you multiply out $$(1+x+x^2+\cdots +x^{10})^4$$ you get terms of the form $$x^{a_1}x^{a_2}x^{a_3}x^{a_4}$$ where $a_1,a_2,a_3,a_4 \in \{0,1,\ldots, 10\}$ and you want  $a_1+a_2+a_3+a_4=10$. Example, you take $x^0$ from the first three terms and $x^{10}$ from the last term would correspond to $a_1=a_2=a_3=0, a_4=10$.
So, how many solutions does this have? This is the stars and bars problem. As others have said, the answer is $$\dbinom{10+4-1}{4-1} = 286$$ Look up the stars and bars problem for more information if this method is easier for you to understand.
A: So we have $$\left(\sum_{i=0}^{10}{x^i}\right)^4=\sum_{l=0}^{10}\sum_{k=0}^{10}\sum_{j=0}^{10}\sum_{i=0}^{10}{x^i x^j x^k x^l}.$$
We want all terms $x^i x^j x^k x^l$ such that $i+j+k+l=10.$ This implies that we simply need the number of partitions of $10,$ times $4!.$ Can you continue from here?
