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Question

Simplify: (n+1)C3(n1)C3=

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Solution

n+1C3 n1C3
(n+1)!(n2)!3!(n1)!(n4)3! as [ nCr=n!(n7)!r!]
13![(n+1)(n)(n1)(n1)(n2)(n3)]
(n1)3![n2+n(n2)(n3)]
(n1)6(n2+n(n22n3n+6))
(n1)6[n2+nn2+5n6]
(n1)6×(6n6)=6(n1)26=(n1)2

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