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Algebra

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2 Jun 2008 12:27:58 IST

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Divisibility

None

Prove that $\frac {n^2!}{n!^n}$ is a natural number

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Bill Gates JR...Don't believe it?..don't ask!!

Scorching goIITian

Joined: 16 Apr 2008

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2 Jun 2008 13:24:09 IST

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Don't know how to solve it mathematically but understanding it would be ..it is an integer because

It is the division of n.n things among n groups each containg n things .

1.7k

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Joined: 31 May 2008

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2 Jun 2008 13:34:13 IST

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n^2! can be written in n lines

1x2x3......xn
(n+1)x(n+2).....2n
...
...
...
................ n^2

clearly, each of these lines are divisible n!. QED!

However, I like risin's combinatorial arguement better. :)

abhishek sinha

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2 Jun 2008 13:39:58 IST

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Yeah, common question with the key in combinatorics !!!

Hari Shankar

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2 Jun 2008 18:55:36 IST

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I was actually looking for an algebraic proof. Not that a combinatorial proof is not welcome, but the algebraic proof will deepen your understanding

One result that is useful here is that the product of any n consecutive numbers is divisible by n!

Now, consists of consecutive numbers i.e. n sub-products of each containing n consecutive numbers

i.e n sub-products each divisible by n!

Hence is a multiple of

In a similar way you can prove that (n!)! is divisible by $(n!)^{(n-1)!}$

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