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!

Since the exponent $2$ sought does not exceed the highest one in $1+x+x^2$, the answer will be the same as the coefficient of $x^2$ in $(1+x+x^2+x^3+\cdots)^{10}=(\frac1{1-x})^{10}$ which is $(-1)^2\binom{-10}2=\binom{11}2=55.$
The combinatorial interpretation is that a contribution to $x^2$ comes from choosing two of the factors $1+x+x^2$, where order is ignored but the two factors may be the same one (in which case it is the term $x^2$ that contributes rather than $x$). That this can be done in $\binom{11}2$ ways can be seen for instance by a stars-and-bars argument.
For other coefficients in $(1+x+x^2)^{10}$ such a simple argument will not work though.
