Thank you for reporting, we will resolve it shortly
Q.
The coefficient of $a^{8} \,b^{4}\, c^{9} \,d^{9}$ in $(a b c+a b d+a c d+b c d)^{10}$ is
Binomial Theorem
Solution:
$a ^{10}\, b ^{10}\, c ^{10} \,d ^{10}\left(\frac{1}{ a }+\frac{1}{ b }+\frac{1}{ c }+\frac{1}{ d }\right)^{10}$
Therefore the required coefficient is equal to the coefficient of
$a ^{-2} \,b ^{-6}\, c ^{-1}\, d ^{-1}$ in $\left(\frac{1}{ a }+\frac{1}{ b }+\frac{1}{ c }+\frac{1}{ d }\right)^{10}$,
which is given by
$\frac{10 !}{2 ! 6 ! 1 ! 1 !}=\frac{10 \times 9 \times 8 \times 7}{2}$
$=2520$