# Finding the coefficient of $x^{25}$ in $(1 + x^3 + x^8)^{10}$?

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

I've tried thinking of this combinatorially, but I couldn't get it to make sense. I've also tried applying some identities, only to lead to dead ends. Any hints?

• Which coefficient? Feb 19 '13 at 9:52
• @rlgordonma Thanks Feb 19 '13 at 9:55
Hint: Can you write $25$ as a sum of eights and threes?
"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 $45×56 = 2520$." Same as asimut with slightly different wording.
• It took me a minute to figure out where the $8$ in ${{8}\choose{3}}$ was coming from. The 3 $x^3$ are being chosen from the remaining $8$ multiplicands. ${10\choose{3}} {7\choose{2}}$ would also be an identical solution. Sep 11 '16 at 23:46
The exponent 25 can arise as $2\cdot 8 + 3\cdot 3 + 5\cdot 0$ only. So you have to count the words of length $10$ consisting of two 8's, three 3's and five 0's. Basic combinatorics gives the result $$\binom{10}{2}\binom{8}{3} = 2520\text{.}$$