What is the exact coefficient of $x^{12}$ in $(2+x^3+x^6)^{10}$? What is the coefficient of $x^{12}$ in $(2+x^3+x^6)^{10}$?
I figure you need to pick $x^3$ 4 times so $C(10,4)$...but what happens with the other numbers/variables???
Can someone explain to me how this is done properly?
Thanks.
EDIT:
$(x + y)^n = C(n,k) \cdot x^{n-k} \cdot y^k$
EX: Find the term for $x^5$ in $(5-2x)^8$
Answer: $C(8,5) \cdot (-2)^5 \cdot 5^3$
How can I use this info to solve a polynomial based question such as the featured?
 Answer: 
To sum up all information provided by everyone (Thanks!!!):
$(C(10,4) * 2^6) + (C(10,2) * 2^8) + (C(10,1) * C(9,2) + 2^7) = 71040 $

You can reach $x^{12}$ in the product from $x^6x^6\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2$, which can be done in $\binom{10}{2}$ ways with a coefficient of $2^8$. 
Or you can reach it with $x^6x^3x^3\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2$, which can be done in $\binom{10}{1}\binom{9}{2}$ ways with a coefficient of $2^7$.
Or you can reach it with $x^3x^3x^3x^3\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2$, which can be done in $\binom{10}{4}$ ways with a coefficient of $2^6$.
$$\begin{align}2^8\binom{10}{2}+2^7\binom{10}{1}\binom{9}{2}+2^6\binom{10}{4}&=64(4\cdot45+2\cdot10\cdot36+210)\\
&=64(1110)\\
&=71040
\end{align}$$
