Question

The coefficient of $x^4$ in the expansion of $(1+x+x^2{)}^{10}$ is

Answer 1

General term of the given expression is given by $\frac{10!}{p!q!r!}{x}^{q+2r}$
Here, $q+2r=4$
For $p=6,q=4,r=0$,
coefficient = $\frac{10!}{6!\times 4!}=210$
For $p=7,q=2,r=1$,
coefficient = $\frac{10!}{7!\times 2!\times 1!}=360$
For $p=8,q=0,r=2$,
coefficient = $\frac{10!}{8!\times 2!}=45$
Therefore, sum = 615