Too many variables?

$\displaystyle S = \sum_{k=8}^{35} k(k-1)(k-2)(k-3)(k-4)(k-5)(k-6)$

Given that the sum above can be expressed as $a! \times \left( \dfrac{b \times c \times d \times e \times f \times g \times h \times i}{j!} - 1 \right)$

where $$a,b,c,d,e,f,g,h,i$$ are all positive integers with $$j$$ is the cube of a prime number and $$b,c,d,e,f,g,h,i$$ are consecutive integers.

Find $$a + b + c + d + e + f +g + h + i + j$$.

Bonus: Generalize the sum,
$$\displaystyle \sum_{m}^{n} (k)(k-1)(k-2)\cdots(k-(r-1))$$ such that $$m > r$$.

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