`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!`

Options

A. `2!3!4!5!6!`
B. `(5!)^(4)`
C. `(10!)^(2)`
D. `10!11!`

Answer

A, B, C, D

Explanation

`(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!`