How to find the coefficient of $x^2y^3z^6$ in the expression $(x^2+y+z)^{10}$? How can find the coefficient of $x^2y^3z^6$  in the expression $(x^2+y+z)^{10}$? 
I know this involves the binomial theorem, but I can't figure out how to solve it. 
 A: We can apply the binomial theorem twice in order to find the coeffcient. It is convenient to use the coefficient of operator $[x^k]$ to denote the coefficient of $x^k$ of an expression.

We obtain
  \begin{align*}
[x^2y^3z^6](x^2+y+z)^{10}&=[x^2y^3z^6]\sum_{k=0}^{10}\binom{10}{k}x^{2k}(y+z)^{10-k}\tag{1}\\
&=[y^3z^6]\binom{10}{1}(y+z)^9\tag{2}\\
&=10[y^3z^6]\sum_{k=0}^9\binom{9}{k}y^kz^{9-k}\tag{3}\\
&=10\binom{9}{3}\tag{4}\\
&=840
\end{align*}

Comment:


*

*In (1) we apply the binomial theorem once to $(x^2+(y+z))^{10}$.

*In (2) we select the coefficient of $x^2$ which is given when the index $k=1$.

*In (3) apply the binomial theorem again.

*In (4) we select the coefficient of $y^3z^6$ corresponding to the index $k=3$.
A: Hint:
The formula to find the coefficient of $x_1^{r_1}x_2^{r^2}\dots x_k^{r_k}$ in $(x_1+x_2+\dots+x_k)^n$ is:
$$\frac{n!}{r_1!r_2!\dots r_k!}$$
Here you take $x^2 =p$ say, then the problem reduced to finding coeff of  $py^3z^6$  in the expression $(p+y+z)^{10}$.
Ans: $$\frac{10!}{1!3! 6!}$$
A: Think of it this way may be intuitively helpful. How many different $10$-tuples can you make out of $x^{2}, y, y, y, z, z, z, z, z, z$? The number of each "character" corresponds to each of the exponent given; using the notion of tuple is due to the commutative multiplication law. Now treat the same letters as if they were different (something like $y_{1},y_{2},y_{3},...$, which you consider mentally.); then you have got $10!$ tuples certainly. But this is way too many! We have to tailor it by observing that this number is actually $3! \times 6!$ (due to the duplicates of $y$ and $z$, respectively.) times the required number! So the required number is
$$
\frac{10!}{3!6!}.
$$
