Coefficient of $x^{25}$

Question

How would one go about finding the coefficient of $x^{25}$ in $(1+x^3 + x^8)^{10}$?

I've tried seeing if I can make the polynomial look like an identity, but I can't see to.

Answer 1

Hint. By expanding two binomials, we have that $$(1+x^3 + x^8)^{10}=(1+x^3(1 + x^5))^{10}=\sum_{k=0}^{10}\binom{10}{k}x^{3k}(1+x^5)^k =\sum_{k=0}^{10}\sum_{j=0}^{k}\binom{10}{k}\binom{k}{j}x^{3k+5j}.$$ Now note that $25=3\cdot 5+5\cdot 2$. Are there any other ways to write $25$ as $3k+5j$ with $0\leq j\leq k\leq 10$? What is the coefficient of $x^{25}$?

Answer 2

Here's a hint.

Can you write $25$ as a sum of $8$'s and $3$'s?

Answer 3

When multiplying out the product, the term $x^{25}$ will arise a number of times, depending on how many times one can write $25$ as a sum of $3$'s and $8$'s where the order of summation matters. It's clear that we need at least one $8$ to write $25$ like this, since $3$ doesn't divide $25$. For the same reason, we see that we actually need exactly two $8$'s, and then $(25-16)/3=3$ $3$'s. So how many ways are there of permuting the terms on the left hand side of $8 + 8 + 3 + 3 + 3 = 25$? Well, exactly ${5}\choose{2}$ $= 10$ ways since the order of the $8$'s and $3$'s is fixed once the $8$'s have been placed. But now, since there are $10$ factors of the polynomial $1 + x^3 + x^8$, we need to multiply $10$ by the number of ways of choosing $5$ factors out of the $10$ factors. So the result will be ${10}\choose{5}$$\cdot 10 = 2520$.

Answer 4

As for the binomial development, there is also the trinomial development.

$$(a+b+c)^n=\sum\limits_{i+j+k=n} \binom{n}{i,j,k}a^ib^jc^k$$

with $\displaystyle \binom{n}{i,j,k} =\dfrac{n!}{i!j!k!}$

In our case $a=1$, $b=x^3$ and $c=x^8$, thus

$$(1+x^3+x^8)^{10}=\sum\limits_{i+j+k=10} \binom{10}{i,j,k}1^ix^{3j}x^{8k}=\sum\limits_{i+j+k=10} \binom{10}{i,j,k}x^{3j+8k}$$

We want the coefficient of $x^{25}$ so we need $3j+8k=25\iff j=3,k=2$ and this forces $i=5$.

The coefficient is then $\displaystyle\binom{10}{5,3,2}=\dfrac{10!}{5!3!2!}=\dfrac{10.9.8.7.6.5.4.3.2.1}{(5.4.3.2.1).(3.2.1).(2.1)}=5.9.8.7=2520$