answer is in the question itself

$\displaystyle (abc+abd+acd+bcd)^{10}$

$\displaystyle = (abcd) ^{10} \times (\frac{1} {a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d})^{10}$

$\text{ }$

We can see that for getting $\displaystyle a^8 b^4 (cd)^9$ , the power of $\displaystyle \frac{1}{a}$ has to be $\displaystyle 2$ , that of $\displaystyle \frac{1}{b}$ to be $\displaystyle 6$ and that of $\displaystyle \frac{1}{c} , \frac{1}{d}$ to be $\displaystyle 1$ each.

$\text{ }$

So, the coefficient of $\displaystyle a^8 b^4 (cd)^9$ will be:

$\displaystyle \Rightarrow \frac{10!}{2! \cdot 6! \cdot 1! \cdot 1!}$

$\displaystyle \Rightarrow \boxed{2520}$

Therefore, the sum of the digits is $2+5+2+0 = \boxed{9}$