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`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!` |
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Answer» Correct Answer - 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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