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

The Correct Option is C

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 \( \beta \in \mathbb{N} \)