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!
 A: Method $\#1:$
Following your process, $$(1+x+x^2)^{10}=\left(\frac{1-x^3}{1-x}\right)^{10} =(1-x^3)^{10}(1-x)^{-10}$$
$$=\left[1-\binom{10}1x^3+\binom{10}2x^6-\cdots\right]\left[1-10(-x)+\frac{-10(-10-1)}2(-x)^2+\cdots\right]$$
Clearly the coefficient of $x^2$ is $\displaystyle \frac{-10(-10-1)}2=55$
Method $\#2:$
For a clearer version of user$125053$'s answer
$$(1+x+x^2)^{10}=\left[(1+x)+x^2\right]^{10}$$
$$=\binom{10}0(1+x)^{10}+\binom{10}1(1+x)^9(x^2)^1+\binom{10}1(1+x)^8(x^2)^2\cdots$$
So, the required coefficient of $x^2$ is
the coefficient of $x^2$ in $\displaystyle\binom{10}0(1+x)^{10}+$ the coefficient of $x^0$ in $\displaystyle\binom{10}1(1+x)^9$
$\displaystyle=\binom{10}0\cdot\binom{10}2+\binom{10}1\cdot\binom90 =55 $ again
Method $\#3:$
$$(1+x+x^2)^{10}=\left[1+x(1+x)\right]^{10}$$
$$=1+\binom{10}1x(1+x)+\binom{10}2\{x(1+x)\}^2+\binom{10}3\{x(1+x)\}^3+\cdots$$
$$=1+\binom{10}1(x+x^2)+\binom{10}2\{x^2(1+x)^2\}+\binom{10}3\{x(1+x)\}^3+\cdots$$
So, the required coefficient of $x^2$ is  $\displaystyle\binom{10}1+\binom{10}2=55$
Method $\#4:$
Using Multinomial Theorem,
the general term of the expansion of $\displaystyle(1+x+x^2)^{10}$
is $\displaystyle\frac{10!}{a!b!c!}(1)^a(x)^b(x^2)^c=\frac{10!}{a!b!c!}x^{b+2c}$  where $a,b,c\ge0$ and $a+b+c=10$
Now we need $b+2c=2\iff2c=2-b\le2$ as $b\ge0$
$\displaystyle\implies b\le1\implies b=0,1$
If $b=0,\cdots$
If $b=1,\cdots$
A: So think about all the different ways you can get $x^2$ when you do expand it.  Either you have a single $x^2$ being multiplied by $9$ other $1$'s, or $x$ multiplied my another $x$ multiplied by $8$ other $1$'s.  It shouldn't be too hard to count all the ways.
A: As you noted:
$$
    [x^2]\left(1+x+x^2\right)^{10} = [x^2] \frac{(1-x^3)^{10}}{(1-x)^{10}}
$$
Because $x^3$ will only contribute to terms $x^3$ and higher order
$$
  [x^2] \frac{(1-x^3)^{10}}{(1-x)^{10}} = [x^2] \frac{1}{(1-x)^{10}} = [x^2] \frac{1}{(1-x)^{10}}
$$
Now using $n \cdot [x^n] g(x) = [x^{n-1}] g^\prime(x)$ with $n=2$ and $g(x) = \frac{1}{9}(1-x)^{-9}$:
$$
  [x^2] \frac{1}{(1-x)^{10}} = 3 [x^{3}] \frac{1}{9 (1-x)^9} = 3 \cdot 4 [x^4] \frac{1}{9 \cdot 8 (1-x)^8} = \frac{(11)!}{2} \cdot [x^{11}] \frac{1}{9! \cdot (1-x)} 
$$
Using $[x^n] \left(1-x\right)^{-1} = 1$ for $n\geq 1$ we arrive at:
$$
   [x^2] \frac{1}{(1-x)^{10}} = \frac{(11)!}{2 \cdot 9!} = \frac{11 \cdot 10}{2} = 55
$$
A: 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
$\displaystyle \binom{10}{9} \binom{9}{0} + \binom{10}{10} \binom{10}{2} = 10 \cdot 1 + 1 \cdot 45 = 55$
A: 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.
