<s>[INST] coefficient of $x^2$, in $(1+x+x^2)^{10}$ 
How  to find coefficient of $x^2$, in $(1+x+x^2)^{10}$, without actually expanding it?

I think the fact $\dfrac{1-x^3}{1-x}=1+x+x^2$ may help.
But can't use it!
 [/INST] Using the binomial theorem,
$(1 + x + x^2)^{10} = \displaystyle \sum_{j=0}^{10} \binom{10}{j} (1 + x)^j (x^2)^{10 - j}$
Using the binomial theorem again,
$\begin{align*}
(1 + x + x^2)^{10} & = \displaystyle \sum_{j=0}^{10} \binom{10}{j} \left( \sum_{k=0}^j \binom{j}{k} x^k \right) (x^2)^{10 - j} \\
                   & = \sum_{j=0}^{10} \sum_{k=0}^j \binom{10}{j} \binom{j}{k} x^{20 + k - 2j} 
\end{align*}$
Now, by parity, $k$ must be even. There are only two possbilities: $k = 0, j = 9$ and $k = 2, j = 10$. Thus, the coefficient is
$\displaystyle \binom{10}{9} \binom{9}{0} + \binom{10}{10} \binom{10}{2} = 10 \cdot 1 + 1 \cdot 45 = 55$
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