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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?

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    $\begingroup$ Which coefficient? $\endgroup$
    – Ron Gordon
    Feb 19 '13 at 9:52
  • $\begingroup$ @rlgordonma Thanks $\endgroup$ Feb 19 '13 at 9:55
  • $\begingroup$ This may help you. $\endgroup$
    – user10676
    Feb 19 '13 at 12:06
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Hint: Can you write $25$ as a sum of eights and threes?

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  • $\begingroup$ How does the 10 play a role? $\endgroup$ Feb 19 '13 at 9:57
  • $\begingroup$ @AlanH The 10 tells you how many choices you have as to where you can "place" each 8 and 3. Azimut has provided a more full solution. $\endgroup$
    – JSchlather
    Feb 19 '13 at 10:17
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"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.

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    $\begingroup$ 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. $\endgroup$ Sep 11 '16 at 23:46
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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 $$\binom{10}{2}\binom{8}{3} = 2520\text{.}$$

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