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Question

If P=n(n212)(n222)(n232)...(n2r2),n>r,nϵ N, then P is divisible by

A
(2r+2)!
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B
(2r1)!
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C
(2r+1)!
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D
none of these
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Solution

The correct options are
B (2r1)!
D (2r+1)!
P=n(n21)(n222)...(n2r2)
P=n(n1)(n+1)(n2)(n+2)...(nr)(n+r)
Rearranging the terms we get,
P=(nr)(nr+1)...(n+r1)(n+r)
This is a product of 2r+1 consecutive terms. Hence it must be divisible by (2r+1)!
Hence, options B and C are correct.

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