Question: using theorem k (m) (n) = (m + n) r=0 (r) (k-r) (k) [ please consider (n) as combinations] (r) n 1.16) show that

using theorem k (m) (n) = (m + n) r=0 (r) (k-r)

using theorem k (m) (n) = (m + n) r=0 (r) (k-r) (k) [ please consider (n) as combinations] (r) n 1.16) show that (n) = (2n) r = 0 (r) (n)

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