Question

If the coefficient of $a^8{b}^4{c}^9{d}^9$ in the expansion of $(abc+abd+acd+bcd{)}^{10}$ is $N$, then what is the sum of digits of $N$?

Answer 1

The general term of ${(abc+abd+acd+bcd)}^{10}$ is $\frac{10!}{x!y!z!p!}{\left(abc\right)}^x{\left(abd\right)}^y{\left(acd\right)}^z{\left(bcd\right)}^p=\frac{10!}{x!y!z!p!}{a}^{(x+y+z)}{b}^{(x+y+p)}{c}^{(x+z+p)}{d}^{(y+z+p)}$

We need coefficient of $a^8{b}^4{c}^9{d}^9$

$\Rightarrow x+y+z=8,x+y+p=4,x+z+p=9,y+z+p=9$

By solving , we get $x=1,y=1,z=6,p=2$

therefore the coefficient is $\frac{10!}{1!1!6!2!}=2520$

The sum of digits is $2+5+2+0=9$