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Prove by permutation of otherwise (n 2 )! / (n!) n is an integer (n ∈ I).

Prove by permutation of otherwise (n2)! / (n!)n  is an integer (n ∈ I).

Grade:upto college level

1 Answers

Navjyot Kalra
askIITians Faculty 654 Points
9 years ago
Hello Student,
Please find the answer to your question
Let there be n sets of different objects each set containing n identical objects
[eg (1, 1, 1 . . . . . . . . . . . (n times) ), (2, 2, 2 . . . . . . . . . . . . . . . 2 (n times)) . . . . . . . . . . . . . . . (n, n, n . . . . . . . n) n times) )]
Then the no. of ways in which these n x n = n2 objects can be arranged in a row
= (n2)! / n! n! . . . . . . . n! = (n2)! / (n!)n
But these number of ways should be a natural number
Hence (n2)! / (n!)n is an integer. (n ∈ I)

Thanks
navjot kalra
askIITians Faculty

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