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

Question

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

Accepted Answer

$$\begin{aligned}
& \alpha=\frac{(4 !) !}{(4 !)^{3 !}}, \beta=\frac{(5 !) !}{(5 !)^{4 !}} \
& \alpha=\frac{(24) !}{(4 !)^6}, \beta=\frac{(120) !}{(5 !)^{24}}
\end{aligned}$$

Let 24 distinct objects are divided into 6 groups of 4 objects in each group.

No. of ways of formation of group $$=\frac{24 !}{(4 !)^6 .6 !} \in \mathrm{N}$$
Similarly,

Let 120 distinct objects are divided into 24 groups of 5 objects in each group.

No. of ways of formation of groups

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

Solutions · 1