Key Concepts and Formulas

• Factorial: $n! = n \times (n-1) \times (n-2) \times \dots \times 2 \times 1$.
• Number of ways to divide $mn$ distinct objects into $n$ groups of $m$ objects each: $\frac{(mn)!}{(m!)^n}$.
• Number of ways to divide $mn$ distinct objects into $n$ groups of $m$ objects each, where the order of the groups does not matter: $\frac{(mn)!}{(m!)^n n!}$.

Step-by-Step Solution

Step 1: Simplify $\alpha$ and $\beta$

First, we calculate the values of the factorials inside $\alpha$ and $\beta$. $4! = 4 \times 3 \times 2 \times 1 = 24$ $5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$

Substituting these values into the expressions for $\alpha$ and $\beta$, we get: $\alpha = \frac{(4!)!}{(4!)^{3!}} = \frac{(24)!}{(24)^{6}}$ (Since $3! = 3 \times 2 \times 1 = 6$) $\beta = \frac{(5!)!}{(5!)^{4!}} = \frac{(120)!}{(120)^{24}}$ (Since $4! = 24$)

Step 2: Analyze $\alpha$

We have $\alpha = \frac{(24)!}{(24)^6}$. We can write $(24)^6 = 24 \times 24 \times 24 \times 24 \times 24 \times 24$. Also, $(24)! = 24 \times 23 \times 22 \times \dots \times 2 \times 1$. So, $\alpha = \frac{24 \times 23 \times 22 \times \dots \times 1}{24 \times 24 \times 24 \times 24 \times 24 \times 24} = \frac{23 \times 22 \times \dots \times 1}{24 \times 24 \times 24 \times 24 \times 24}$. Since the numerator is a product of integers and the denominator is a product of integers, $\alpha$ will be a rational number. We need to determine if it is an integer.

Consider dividing 24 distinct objects into 6 groups, where each group has 4 objects. The number of ways to do this is NOT what we're looking for, as the order of objects within each group MATTERS (permutations). Instead, let's think of $\alpha = \frac{24!}{24^6} = \frac{24!}{24 \cdot 24 \cdot 24 \cdot 24 \cdot 24 \cdot 24}$. $(24)! = 1 \cdot 2 \cdot 3 \cdot ... \cdot 24$. In this expansion, we have factors of 24, and we are dividing by 24 six times. Since 24! will have factors of $24^6$, it's likely that the result is an integer.

However, let's think about it differently. Consider dividing 24 distinct objects into groups of size 1. Then $(24)!$ represents all possible permutations of these objects. Now, consider $\frac{24!}{24^6}$. This is NOT directly related to a counting problem where we know the answer must be an integer.

Let's try to determine if $\alpha$ is an integer by examining the prime factorization of the numerator and denominator. $24 = 2^3 \cdot 3$. Thus, $24^6 = (2^3 \cdot 3)^6 = 2^{18} \cdot 3^6$. In $24!$, the exponent of 2 is $\lfloor \frac{24}{2} \rfloor + \lfloor \frac{24}{4} \rfloor + \lfloor \frac{24}{8} \rfloor + \lfloor \frac{24}{16} \rfloor = 12 + 6 + 3 + 1 = 22$. In $24!$, the exponent of 3 is $\lfloor \frac{24}{3} \rfloor + \lfloor \frac{24}{9} \rfloor = 8 + 2 = 10$. Thus, $24! = 2^{22} \cdot 3^{10} \cdot \dots$. Since $24^6 = 2^{18} \cdot 3^6$, $\frac{24!}{24^6} = \frac{2^{22} \cdot 3^{10} \cdot \dots}{2^{18} \cdot 3^6} = 2^4 \cdot 3^4 \cdot (\text{other primes})$. Since the exponents of the primes in the numerator are greater than or equal to the exponents of the primes in the denominator, $\alpha$ is an integer. So, $\alpha \in \mathbb{N}$.

Step 3: Analyze $\beta$

We have $\beta = \frac{(120)!}{(120)^{24}}$. We can write $(120)^{24} = 120 \times 120 \times \dots \times 120$ (24 times). Also, $(120)! = 120 \times 119 \times 118 \times \dots \times 2 \times 1$. So, $\beta = \frac{120 \times 119 \times 118 \times \dots \times 1}{120 \times 120 \times \dots \times 120} = \frac{119 \times 118 \times \dots \times 1}{120 \times 120 \times \dots \times 120}$ (23 times).

Consider dividing 120 distinct objects into 24 groups, where each group has 5 objects. The number of ways to do this is NOT what we're looking for, as the order of objects within each group MATTERS (permutations).

Let's think of $\beta = \frac{120!}{120^{24}}$. $(120)! = 1 \cdot 2 \cdot 3 \cdot ... \cdot 120$. In this expansion, we have factors of 120, and we are dividing by 120 twenty-four times. Since 120! will have factors of $120^{24}$, it's likely that the result is an integer.

Let's determine if $\beta$ is an integer by examining the prime factorization of the numerator and denominator. $120 = 2^3 \cdot 3 \cdot 5$. Thus, $120^{24} = (2^3 \cdot 3 \cdot 5)^{24} = 2^{72} \cdot 3^{24} \cdot 5^{24}$. In $120!$, the exponent of 2 is $\lfloor \frac{120}{2} \rfloor + \lfloor \frac{120}{4} \rfloor + \lfloor \frac{120}{8} \rfloor + \lfloor \frac{120}{16} \rfloor + \lfloor \frac{120}{32} \rfloor + \lfloor \frac{120}{64} \rfloor = 60 + 30 + 15 + 7 + 3 + 1 = 116$. In $120!$, the exponent of 3 is $\lfloor \frac{120}{3} \rfloor + \lfloor \frac{120}{9} \rfloor + \lfloor \frac{120}{27} \rfloor + \lfloor \frac{120}{81} \rfloor = 40 + 13 + 4 + 1 = 58$. In $120!$, the exponent of 5 is $\lfloor \frac{120}{5} \rfloor + \lfloor \frac{120}{25} \rfloor = 24 + 4 = 28$. Thus, $120! = 2^{116} \cdot 3^{58} \cdot 5^{28} \cdot \dots$. Since $120^{24} = 2^{72} \cdot 3^{24} \cdot 5^{24}$, $\frac{120!}{120^{24}} = \frac{2^{116} \cdot 3^{58} \cdot 5^{28} \cdot \dots}{2^{72} \cdot 3^{24} \cdot 5^{24}} = 2^{44} \cdot 3^{34} \cdot 5^4 \cdot (\text{other primes})$. Since the exponents of the primes in the numerator are greater than or equal to the exponents of the primes in the denominator, $\beta$ is an integer. So, $\beta \in \mathbb{N}$.

Step 4: Final Conclusion

Since $\alpha \in \mathbb{N}$ and $\beta \in \mathbb{N}$, option (A) is the correct answer.

Common Mistakes & Tips

• Don't confuse dividing into groups where order matters within the group versus combinations where order doesn't matter.
• Prime factorization is a powerful tool for determining if a fraction is an integer.
• Carefully calculate the exponents of primes in factorials using Legendre's Formula.

Summary

We simplified the expressions for $\alpha$ and $\beta$ and then used prime factorization to determine whether they are integers. We found that $\alpha$ and $\beta$ are both integers. Therefore, the correct option is (A).

Final Answer

The final answer is \boxed{\alpha \in \mathbf{N} and \beta \in \mathbf{N}}, which corresponds to option (A).