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

To solve this problem, we need to determine whether the expressions \( \alpha \) and \( \beta \) belong to the set of natural numbers \( \mathbb{N} \).

1. Expression for \( \alpha \):  
   We have \(\alpha = \frac{(4!)!}{(4!)^{3!}}\). First, calculate \(4!\): \(4! = 4 \times 3 \times 2 \times 1 = 24\). 
   Now substitute into the expression for \(\alpha\): \(\alpha = \frac{24!}{24^{6}}\). 
   Here, \(24!\) represents the factorial of \(24\), which is the product of all positive integers up to \(24\). 
   The denominator, \(24^{6}\), suggests raising \(24\) to the power \(6\). 
   Since \(24!\) is a factorial, it includes \(24\) as a factor many times, ensuring that \(24^{6}\) will divide \(24!\) without leaving any decimal or fraction, thus \(\alpha \in \mathbb{N}\).
2. Expression for \( \beta \): 
   We have \(\beta = \frac{(5!)!}{(5!)^{4!}}\). Calculate \(5!\): \(5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\). 
   Now substitute into the expression for \(\beta\): \(\beta = \frac{120!}{120^{24}}\). 
   Again, \(120!\) includes \(120\) as a factor repeatedly, more than sufficient for ensuring \(120^{24}\) divides \(120!\) completely, so \(\beta \in \mathbb{N}\).
3. Conclusion: 
   Both \(\alpha\) and \(\beta\) are 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 \( \beta \in \mathbb{N} \)