Find the coefficient of $x^{25}$ in $(1 + x^3 + x^8)^{10}$ 
Find the coefficient of $x^{25}$ in $(1 + x^3 + x^8)^{10}$.

Here is my solution I am looking to see if it is correct or if there is another way to do it, thanks!
The only way to form an $x^{25}$ term is to gather two $x^8$ and three $x^3$ . Since there are ${{10}\choose{2}} =45$ ways to choose two $x^8$ from the $10$ multiplicands and $8$ ways to choose three  ${{8}\choose{3}}= 56$ ways to choose $x^3$ from the remaining $8$multiplicands, the answer is using the product rule  $45×56 = 2520$.

A formal way:
$$(1 + x^3 + x^8)^{10}=\sum_{k=0}^{10}\binom{10}{k}x^{3k}(1+x^5)^k=\sum_{k=0}^{10}\sum_{j=0}^{k}\binom{10}{k}\binom{k}{j}x^{3k+5j}.$$
Now $25=3k+5j$ for $ 0\leq j\leq k\leq 10$ is solved iff $(k,j)=(5,2)$ and
$$[x^{25}](1 + x^3 + x^8)^{10}=\binom{10}{5}\binom{5}{2}=2520.$$
