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11 Sep 2009 00:37:44 IST

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Show that (6!)^5! is a divisor of(6!)!.

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Show that (6!)^5! is a divisor of(6!)!.

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kabi

Blazing goIITian

Joined: 11 Jan 2009

Posts: 575

11 Sep 2009 12:50:47 IST

Like 3 people liked this

this can be done by PnC

let solve it for general

that is To show ( n! )! is divisible by ( n! )^(n-1)!

take any m consective numbers let say - ( a ,a +1 ,a+2 ,..................a +m-1 )

now product of all these numbers = ( a )(a+1)................(a+m-1) = (a +m-1)! / (a-1)!

= m ! * ^a+m-1C_m

_{that mean}

m consective numbers 's product is divisible by m!

as ^a+m-1C_mis an integer

so product of 1 to n integers is divisible by n! .............................( 1st bunch of n integers )

and product of n+1 to 2n is also divisible by n!...............................(2nd bunch of n integers )

......

....

.....

....

.... and product of n! - n+1 to n! is also divisible by n ! .........................((n-1)! bunch of n integers )

product of integers from 1 to n! = ( n ! ) !

if we take n! ( which is diviser of every bunch of n intergers ) we will get = ( n! )^(n-1)!

so ( n! )! is divisible by ( n! )^(n-1)!

thank u

( note - Plz try to remember this result as its very important )

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