IIT-2004

Show that (n²)! is divisible by (n!)^n

6 Answers

2nd Q: Show that (n²)! is divisible by (n!)^(n+1)

Q1 if n^2 objects are distributed into "n" no. of equal groups(named)

No. of ways =(n²!)/(n!)ⁿ

Since no. of ways is always integer qty. that implies Numerator is divisble by Dr..

yeah thats right :)

to be more general
(n!)^k divides (nk)!

anyone for 2?

For the 2nd qsn u may simple use the fact that product of p consecutive integers is divisible by p! .

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