Double Sum and Double Product

Probability Level 4

p=1nm=pn(nm)(mp)\displaystyle \sum^{n}_{p=1} \sum^{n}_{m=p} \dbinom{n}{m} \dbinom{m}{p}

If the value of the above expression is in the form anbna^n-b^n, where aa and bb are prime numbers, find a+ba+b.

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