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 $

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.
