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

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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