If P=n(n2−12)(n2−22)(n2−32)...(n2−r2),n>r,nϵN, then P is divisible by
A
(2r+2)!
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B
(2r−1)!
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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(2r−1)! D(2r+1)! P=n(n2−1)(n2−22)...(n2−r2) P=n(n−1)(n+1)(n−2)(n+2)...(n−r)(n+r) Rearranging the terms we get, P=(n−r)(n−r+1)...(n+r−1)(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.