Too many variables?

$\displaystyle S = \sum_{m}^{n} (k)(k-1)(k-2)......(k-(r-1))$
Multiplying and dividing by $r!(k-r)!$
$\displaystyle S = r! \times \sum_{m}^{n} \dfrac{k!}{r! \times (k-r)!}$
$\displaystyle S = r! \times \sum_{m}^{n} {k \choose r}$
$\displaystyle S = r! \times \left ( \sum_{r}^{n}{k \choose r} - \sum_{r}^{m-1}{k \choose r}\right)$

Using the Hockey stick identity ,

$\displaystyle \sum_{r}^{n} {k \choose r} = {n + 1 \choose r+1}$

$\displaystyle \therefore S = r! \times \left( {n+1 \choose r+1} - {m \choose r+1} \right)$

For the given sum,
$n = 35$ , $m = 8$ , $r = 7$

$\displaystyle \therefore S = 7! \times \left( {35 + 1 \choose 7 + 1} - {8 \choose 7 + 1} \right)$
$\displaystyle \therefore S = 7! \times \left( {36 \choose 8} - {8 \choose 8} \right)$
$\displaystyle \therefore S = 7! \times \left( \dfrac{29 \times 30 \times 31 \times 32 \times 33 \times 34 \times 35 \times 36}{8!} - 1 \right)$

Comparing,
$a =7, b = 29, c = 30, d = 31, e = 32, f = 33, g = 34, h = 35, i = 36 , j = 8$

$\displaystyle \therefore a + b + c + d + e + f + g + h + i + j = 7 + 29 + 30 + 31 + 32 + 33 +34 + 35 + 36 + 8 = 275$

It is simply a telescoping series ! . Multiply and divide by 8 and write 8 in the numerator as (k+1)- (k-7) . its done!