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# Counting and Probability Problem

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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) of positive integers, where  1 ≤ a,b,c ≤ 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.

I'd really appreciate a full explanation instead of just an answer thanks! :)

Sep 24, 2021