# 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 $45×56 = 2520$.

However, I would write a little more justification for the sentence:

The only way to form an $x^{25}$ term is to gather two $x^8$ and three $x^3$

How do you know this is true?

• through $2\cdot 8 + 3\cdot 3 + 5\cdot 0$? – fr14 Feb 20 '18 at 16:55
• @fr14 Sure, but how do you know that this is the only way to get $x^{25}$? How do you know there isn't another possibility? – 5xum Feb 20 '18 at 16:55
• I am not sure!. – fr14 Feb 20 '18 at 16:57
• @fr14 Then I suggest you give it some thought. – 5xum Feb 20 '18 at 16:57
• is there another possibility? – fr14 Feb 20 '18 at 16:59

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 $$[x^{25}](1 + x^3 + x^8)^{10}=\binom{10}{5}\binom{5}{2}=2520.$$

Expressions like above can be evaluated by the multinomial Theorem:

$(a+b+c)^n = \sum_{k,l,m;k+l+m=n} \frac{n!}{k!l!m!}a^kb^lc^m$.

This sum means you sum over all possible $k,l,m$ with the condition that $k+l+m=n$.

Now you set $n=10,a=1,b=x^3,c=x^8$ and you obtain now that the $x^{25}$ Terms are obtained when $3l+8m=25$ for integer numbers $l,m$. This equation is satisfied only for $m=2,l=3$. Thus you obtain $k = 10-2-3=5$. It is $a^kb^lc^m = x^{25}$ and the resulting binomial coefficient is $\frac{10!}{5!2!3!} = 2520$. With above formula you obtain that the factor is $2520$ and that is correct!