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P=n(n^(2)-1)(n^(2)-4)(n^(2)-9)…(n^(2)-10...

`P=n(n^(2)-1)(n^(2)-4)(n^(2)-9)…(n^(2)-100)` is always divisible by , `(n in I)`

A

`2!3!4!5!6!`

B

`(5!)^(4)`

C

`(10!)^(2)`

D

`10!11!`

Text Solution

Verified by Experts

The correct Answer is:
A, B, C, D
`(a,b,c,d)`
We know that product of `r` consecutive integer is divisible by `r!`
We have `P=(n-10)(n-9)(n-8)…(n+10)` is produt of `21` consecutive integers
Which is divisible by `21!`
`[(n-10)(n-9)(n-8)...n][(n+1)(n+2)(n+3)...(n+10)]`
Which is divisible by `11! 10!`
`[(n-10)...(n-6)]xx[(n-5)...(n-1)]xx[n(n+1)...(n+4)]xx[(n+5)...(n+9)]xx(n+10)`
which is divisible by `(5!)^(4)`.
Also it can be shown that `P` is divisible by `2!3!4!5!6!`
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