Prove the relations
(0n)(pn)+(1n)(p−1n−1)+(2n)(p−2n−2)+⋯+(pn)(0n−p)=2p(pn),
(0n)(pn)−(1n)(p−1n−1)+(2n)(p−2n−2)−⋯+(−1)p(pn)(0n−p)=0.
(Consider the subsets of p elements of [1,n] which contain a given subset of k elements (0⩽k⩽p), and use Exercise 2 for the second formula.)