A) \[(2r+2)!\]
B) \[(2r+4)!\]
C) \[(2r+1)!\]
D) None of these
Correct Answer: C
Solution :
[c] \[P=n(n+1)(n-1)(n+2)(n-2)...(n+r)(n-r).\] \[=\{n(n+1)(n+2)...(n+r)\}\{(n-1)(n-2)...(n-r)\}\] \[=(n+r)(n+r-1)...(n+1)(n)(n-1)...(n-r)\] Clearly P is product of \[(2r+1)\] consecutive integers, so divisible by \[(2r+1)!\]You need to login to perform this action.
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