In the exapnsion of $(1+x+x^3+x^4)^{10}$, find the coefficient of $x^4$

Question

In the exapnsion of $(1+x+x^3+x^4)^{10}$, find the coefficient of $x^4$.

What's the strategy to approach such problems. Writing expansion seems tedious here.

Answer 1

$$=(1+x^3)^{10}(1+x)^{10}$$

So, required coefficient of $x^4$ will be $$\binom{10}0\binom{10}4+\binom{10}1\binom{10}1$$

Answer 2

Referring to Jack Crawford's comment above, there are three possible ways to get $x^4$: $$\underbrace{1\times1\times\cdots\times 1}_{9}\times x^4\\ \underbrace{1\times1\times\cdots\times 1}_{8}\times x\times x^3\\ \underbrace{1\times1\times\cdots\times 1}_{6}\times x\times x\times x \times x$$ The first is the combination: ${10\choose 1}=10$.

The second is the permutation: $P(10,2)=\frac{10!}{8!}=90$.

The third is again combination: ${10\choose 4}=210$.

Hence: $10+90+210=310$.

Answer 3

You can use the multinomial expansion: $$ (a+b+c+d)^{10}=\sum_{p+q+r+s=10}\binom{10}{p\ q\ r\ s}a^pb^qc^rd^s $$ where $$ \binom{10}{p\ q\ r\ s}=\frac{10!}{p!q!r!s!} $$ With $a=1$, $b=x$, $c=x^3$, $d=x^4$, we have $$ a^pb^qc^rd^s=x^{q+3r+4s} $$ We get $q+3r+4s=4$ if and only if $q=4$, $r=0$, $s=0$ or $q=1$, $r=1$, $s=0$ or $q=0$, $r=0$, $s=1$. Thus the coefficient is $$ \binom{10}{6\ 4\ 0\ 0}+\binom{10}{8\ 1\ 1\ 0}+\binom{10}{9\ 0\ 0\ 1} =\frac{10!}{6!4!}+\frac{10!}{8!1!1!}+\frac{10!}{9!1!}=210+90+10=310 $$