Let $\alpha =\frac{(4!)!}{(4!{)}^{3!}}$ and $\beta =\frac{(5!)!}{(5!{)}^{4!}}$. Then :

Given: $\mathrm{\alpha }=\frac{(4!)!}{(4!{)}^{3!}}, \mathrm{\beta }=\frac{(5!)!}{(5!{)}^{4!}}$

$\Rightarrow \mathrm{\alpha }=\frac{\left(24\right)!}{(4!{)}^6}, \mathrm{\beta }=\frac{\left(120\right)!}{(5!{)}^{24}}$

If $24$ distinct objects are divided into $6$ groups of $4$ objects in each group, then the number of ways of formation of group is$=\frac{24!}{(4!{)}^6.6!}\in \mathrm{N}$

Similarly, let $120$ distinct objects are divided into $24$ groups of $5$ objects in each group.

So, the number of ways of formation of groups

$=\frac{\left(120\right)!}{(5!{)}^{24}·24!}\in \mathrm{N}$