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

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Discussion Response Post to: Show that (6!)^5! is a divisor of(6!)!.

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kabi (1665)

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total posts: 575
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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 )

The time u guys take to find the derivative of a function or for finding the equilibrium constant of a reaction or for finding the angle of dispersion of prism or for standing from ur seat to congratulate our team after their win almost in that time one kid die because of poverty.

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