Question

prove that (n!) ! is divisible by (n!) ^ (n-1)!

Answer 1

Rewrite P=(n!)!(n!)(n−1)!P=(n!)!(n!)(n−1)! as (n!)!(n!)(n!)⋯(n!)(n!)!(n!)(n!)⋯(n!)

Number of n!n!s in the denominator is (n−1)!(n−1)!

Now, (n−1)!×n=n!(n−1)!×n=n!. PP is the number of ways of arranging n!n! things in which nn things are of type 1, other nnthings are of type 2 and so on. Hence P is an integer.

Using a different combinatorial interpretation we can even say that,

(n!)!(n!)(n−1)!(a1)!(a2)!....⋯(ar)!(n!)!(n!)(n−1)!(a1)!(a2)!....⋯(ar)!

is also an integer where aiai are non-negative integers and a1+a2+a3⋯+ar=(n−1)!a1+a2+a3⋯+ar=(n−1)!. Try to figure out how you will prove this.

Answer 2

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