Question

Using permutation or otherwise, prove that `(n^2)!/(n!)^n is an integer, where n is a positive integer. (JEE-2004]

Answer 1

Here, `n^(2)` objects are distributed in n groups, each group containing n identical objects.
`therefore` Number of arrangements
`= ""^(n^(2))C_(n) * ""^(n^(2) - n)C_(n) * ""^(n^(2) - 2n)C_(n) * ""^(n^(2) - 3n)C_(n) * ""^(n^(2) - 2n)C_(n) ... ""^(n)C_(n)`
`= ((n^(2))!)/(n!(n^(2) - n)!).((n^(2) - n)!)/(n!(n^(2) - 2n)!) ... (n!)/(n!*1) = ((n^(2))!)/((n!)^(n))`
`rArr` Integer (as number of arrangements has to be integer).