Finding the Coefficient of X^9 in (1+x^3+X^8)^10

Question

This is solved by the following approach

e1 takes values 0 3 8

e2 takes values 0 3 8

.. ..

.. .

and finally it is said that we get 9 when we take ai=3.And the answer become 10c3. Can someone explain this

Answer 1

Hint: Solve for integer solutions to $9 = 8a + 3b + 0c $ subject to $a+b+c = 10$.

Answer 2

Every term in the expansion of $(1+x^3+x^8)^{10}$ is of the form $1^a(x^3)^b(x^8)^c$ with $a+b+c = 10$. It is clear that the only way to get $x^9$ as a term is when $a = 7,b=3,c=0$. And the coefficient corresponding to this term is the number of ways you can choose three $x^3$'s from $10$ i.e $\binom{10}{3}$

Answer 3

The Coefficient of $x^9$ in $\displaystyle(1+x^3+x^8)^{10}=[(1+x^3)+x^8]^{10}=(1+x^3)^{10}+\binom{10}1(1+x^3)^9x^8+\cdots+(x^8)^{10}$

$=$ the Coefficient of $x^9$ in $\displaystyle (1+x^3)^{10}+10x^8(1+x^3)^9$

Now, the Coefficient of $x^9$ in $\displaystyle(1+x^3)^{10}$ is $\displaystyle\binom{10}3$

and the Coefficient of $x^9$ in $\displaystyle10x^8(1+x^3)^9$ will be $10\cdot[$ the Coefficient of $x$ in $(1+x^3)^9]$ which is $0$