Question

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

• A. $\alpha \in \mathbb{N}$ and $\beta \notin \mathbb{N}$
• B. $\alpha \notin \mathbb{N}$ and $\beta \in \mathbb{N}$
• C. $\alpha \in \mathbb{N}$ and $\beta \in \mathbb{N}$
• D. $\alpha \notin \mathbb{N}$ and $\beta \notin \mathbb{N}$

Answer 1

Answer: (C)

$\alpha \in \mathbb{N}$ and $\beta \in \mathbb{N}$

Answer 2

Given:
$\alpha = \frac{(4!)^6}{(4!)^6 \cdot 6!} = \frac{24!}{(4!)^6 \cdot 6!}, \quad \beta = \frac{(5!)^{24}}{(5!)^{24} \cdot 24!} = \frac{120!}{(5!)^{24} \cdot 24!}.$

Analyzing $\alpha$:
Consider dividing 24 distinct objects into 6 groups of 4 objects each. The number of ways to form these groups is given by:
$\alpha = \frac{24!}{(4!)^6 \cdot 6!}.$

Since this is a valid combinatorial expression representing the number of ways to arrange groups, $\alpha \in \mathbb{N}$ (i.e., it is a natural number).

Analyzing $\beta$:
Consider dividing 120 distinct objects into 24 groups of 5 objects each. The number of ways to form these groups is given by:
$\beta = \frac{120!}{(5!)^{24} \cdot 24!}.$

This is also a valid combinatorial expression, implying that $\beta \in \mathbb{N}$.

Conclusion:
Therefore, both $\alpha$ and $\beta$ are natural numbers.

The Correct answer is: $\alpha \in \mathbb{N}$ and $\beta \in \mathbb{N}$