Well I could prove $\frac{(mn)!}{(n!)^{m}{m!}}$ to be an integer by considering there to be m×n different balls and grouping them into m groups consisting of n balls each. But I could not solve this problem using the same logic.
2026-03-30 06:10:15.1774851015
Prove that $\frac{(mn)!}{(n!)^{m+1}}$ is an integer.
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This isn't true. Consider $n=3, m=2$. Then $$ \frac{6!}{3!^3}=\frac{10}3 $$ is not an integer.