JEE MAIN - Mathematics (2024 - 27th January Evening Shift - No. 16)
Let $$\alpha=\frac{(4 !) !}{(4 !)^{3 !}}$$ and $$\beta=\frac{(5 !) !}{(5 !)^{4 !}}$$. Then :
$$\alpha \in \mathbf{N}$$ and $$\beta \in \mathbf{N}$$
$$\alpha \in \mathbf{N}$$ and $$\beta \notin \mathbf{N}$$
$$\alpha \notin \mathbf{N}$$ and $$\beta \in \mathbf{N}$$
$$\alpha \notin \mathbf{N}$$ and $$\beta \notin \mathbf{N}$$
Explanation
$$\begin{aligned} & \alpha=\frac{(4 !) !}{(4 !)^{3 !}}, \beta=\frac{(5 !) !}{(5 !)^{4 !}} \\ & \alpha=\frac{(24) !}{(4 !)^6}, \beta=\frac{(120) !}{(5 !)^{24}} \end{aligned}$$
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
$$=\frac{(120) !}{(5 !)^{24} \cdot 24 !} \in \mathrm{N}$$
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