Let n be a positive integer.

(a) Prove that \(n^3 = n + 3n(n - 1) + 6 \binom{n}{3}\)

by counting the number of ordered triples (a,b,c) where \(1 \le a,b,c \le n\) in two different ways.

(b) Prove that \(\binom{n + 2}{3} = (1)(n) + (2)(n - 1) + (3)(n - 2) + \dots + (k)(n - k + 1) + \dots + (n)(1),\)

by counting the number of subsets of \(\{1, 2, 3, \dots, n + 2\}\) containing three different numbers in two different ways.

Guest Dec 27, 2019