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 Let n=2015 The least positive integer k for which kn2n212n222n332n2(n1)2=r!  for some positive integer r is

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2014
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detailed solution

Correct option is D

We can rewrite the given expression as kn2(n−1)(n+1)(n−2)(n+2)(n−3)(n+3)……….(n+n−1)(n−n+1)=r!⇒kn(1)(2)…(n−1)n(n+1)(n+2)…(2n−1)=r!⇒kn(2n−1)!=r!∴ To convert L.H.S. to a factorial, we shall requirek = 2 which will convert it into (2n)!


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