# Sum it

Algebra Level 4

If $\large\displaystyle \sum^{99}_{k=0}\dbinom{99}{k}$ is divided by $\large\displaystyle\sum^{33}_{k=0}\dbinom{99}{3k},$
the resulting number is of the form $$\large{\dfrac{a}{b}} \dot \large {\dfrac{c^d}{(e^f-1)}},$$ where $$\large{a, b, c, e}$$ are prime numbers. Evaluate $$\large{a+b+c+d+e+f-1}$$.

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