Show that . Give an interpretation involving subsets.

Solution: n n Step 1: we need to show Ck Cn-k And also want to give an interpretation involving subsets. Step 2: n 1) By the definition of Ck n n! C k= k! (nk)! Since k= n-(n-k) (by substituting) n! nC k (n(nk))! (nk)! By the commutative property of multiplication nC = n! = C k (nk)!(n(nk))! n-k That means n n Ck Cn-k Hence the proof.