Question

Let the coeffcient of $x^4$ in the expansion of${(1+x+x^2+x^3)}^{11}$ be $K$. Find sum of digits of $K$.

Answer 1

$x^4$ can be achieved in following ways
$\left(x{)}^4\right(x^2{)}^0(x^3{)}^0{1}^7$ or $\left(x{)}^2\right(x^2{)}^1(x^3{)}^0{1}^8$ or $\left(x{)}^1\right(x^2{)}^0(x^3{)}^1{1}^9$ or $\left(x{)}^0\right(x^2{)}^2(x^3{)}^0{1}^9$
Hence the required coefficient will be
$=\frac{11!}{4!.7!}+\frac{11!}{1!.2!.8!}+\frac{11!}{1!.1!9!}+\frac{11!}{2!.9!}$
$=330+495+110+55$
$=550+330+110$
$=990$