Prove by permutation of otherwise (n 2 )! / (n!) n

Question

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

Navjyot Kalra · Accepted Answer

Hello Student,Please find the answer to your questionLet 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!)nBut these number of ways should be a natural numberHence (n2)! / (n!)n is an integer. (n &isin; I)Thanksnavjot kalraaskIITians Faculty

Algebra> Prove by permutation of otherwise (n 2 )!...

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

Amit Saxena , 10 Years ago

Navjyot Kalra

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Algebra> Prove by permutation of otherwise (n 2 )!...

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

Amit Saxena , 10 Years ago

Navjyot Kalra

Provide a better Answer & Earn Cool Goodies

LIVE ONLINE CLASSES

Ask a Doubt

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