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 $
 A: Hint 1: $x^{12} = x^6 x^6 = x^3 x^3  x^3 x^3 = x^3 x^3 x^6$
How many ways to pick $x^6 x^6$? Everything that's not an $x$ term is a multiplier of $2$. This would be $2^8 {10 \choose 2}$ for a total of $10$ elements.
For four $x^3$ there would be $10 \choose 4$ ways to pick. 
Hint 2: For $x^3 x^3 x^6$ there are $10$ ways to choose the $x^6$, $10-1 \choose 2$ ways to pick  $x^3$.
Hopefully you can figure out the final sum of the three cases from here.
A: 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 $2^6$.
$$\begin{align}2^8\binom{10}{2}+2^7\binom{10}{1}\binom{9}{2}+2^6\binom{10}{4}&=64(4\cdot45+2\cdot10\cdot36+210)\\
&=64(1110)\\
&=71040
\end{align}$$
A: Hint 1 Let $y=x^3$. 
Hint 2  The coefficient for $y^4$ in $P(y)$ can be calculated by derivating $P(y)$ four times.
A: In combinatorial terms, you're looking for the number of partitions of $12$ that use only $0, 3$, and $6$ and have $10$ parts.  You then weight these partitions by multinomial coefficients.  There are $3$ such partitions:
\begin{align*}
12 &= 6 + 6 + 0 + \cdots + 0\\
&= 6 + 3 + 3 + 0 + \cdots + 0\\
&= 3 + 3 + 3 + 3 + 0 + \cdots + 0 \, .
\end{align*}
These partitions correspond to the terms
\begin{align*}
x^6 * x^6 * 2^8 &= 2^8 x^{12}\\
x^6 * x^3 * x^3 * 2^7 &= 2^7 x^{12}\\
x^3 * x^3 * x^3 * x^3 * 2^6 &= 2^6 x^{12} \, .
\end{align*}
Now we observe that the number of ways each of these can be chosen is $\binom{10}{8,2}$, $\binom{10}{7,2,1}$, $\binom{10}{6,4}$ ways.  So it looks to me that the coefficient of $x^{12}$ is
$$
\binom{10}{8,2} 2^8 + \binom{10}{7,2,1} 2^7 + \binom{10}{6,4} 2^6 \, .
$$
