In the exapnsion of $(1+x+x^3+x^4)^{10}$, find the coefficient of $x^4$ 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.
 A: $$=(1+x^3)^{10}(1+x)^{10}$$
So, required coefficient of $x^4$ will be $$\binom{10}0\binom{10}4+\binom{10}1\binom{10}1$$
A: 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
$$
\binom{10}{6\ 4\ 0\ 0}+\binom{10}{8\ 1\ 1\ 0}+\binom{10}{9\ 0\ 0\ 1}
=\frac{10!}{6!4!}+\frac{10!}{8!1!1!}+\frac{10!}{9!1!}=210+90+10=310
$$
A: 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: $10+90+210=310$. 
