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.

Try:
$\begin{align*}
1 + x + &x^2 + x^3 + x^4 + x^5 + x^6 + x^7 + x^8 + x^9 + x^{10} \\
  &= \frac{1 - x^{11}}{1- x} \\
(1 + x + &x^2 + x^3 + x^4 + x^5 + x^6 + x^7 + x^8 + x^9 + x^{10})^4 \\
  &= \frac{(1 - x^{11})^4}{(1- x)^4} \\
  &= (1 - 4 x^{11} + 6 x^{22} - 4 x^{33} + x^{44})
          \cdot \sum_{k \ge 0} (-1)^k \binom{-4}{k} x^k \\
[x^{10}]  (1 - &4 x^{11} + 6 x^{22} - 4 x^{33} + x^{44})
          \cdot \sum_{k \ge 0} (-1)^k \binom{-4}{k} x^k \\
  &= [x^{10}] \sum_{k \ge 0} (-1)^k \binom{-4}{k} x^k \\
  &= (-1)^{10} \binom{-4}{10} \\
  &= \binom{10 + 4 - 1}{4 - 1} \\
  &= 286
\end{align*}$
