Let α = ( 4 ! ) ! ( 4 ! ) ^ 3 ! and β = ( 5 ! ) ! ( 5 ! ) ^ 4 ! ....

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

$⇒ \alpha = \dfrac{( 24 ) !}{( 4 ! {)}^{6}}, \beta = \dfrac{( 120 ) !}{( 5 ! {)}^{24}}$

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$= \dfrac{24 !}{( 4 ! {)}^{6} . 6 !} \in N$

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

$= \dfrac{( 120 ) !}{( 5 ! {)}^{24} \cdot 24 !} \in N$