Combinatorial sum!

Probability Level 4

If k=018(1)k(18k)k3+9k2+26k+24=1p\displaystyle \sum_{k=0}^{18} \frac{(-1)^k {18 \choose k}}{k^3 + 9k^2 + 26k + 24} = \frac{1}{p} , find pp.

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