Let α = (4!)! / (4!)^3! and β = (5!)! / (5!)^4!. Then : (1) α ∈ N and β ∉ N (2) α ∉ N and β ∈ N (3)

Q: Why can't I just calculate the values? They look like standard factorials.

Calculating $(24)!$ or $(120)!$ is impossible by hand and unnecessary. The beauty of JEE Advanced is that it tests your ability to recognize structural patterns rather than brute-force arithmetic. If you see a massive factorial expression, always look for a combinatorial interpretation first.

Q: How do I know if the result is a natural number?

In combinatorics, if an expression represents the number of ways to perform a physical task—like dividing objects into groups—it must be an integer. Since we proved that our expression represents $n!$ times the number of ways to partition a set, and both $n!$ and the number of ways are integers, the result must be a natural number.

Q: What if the denominator was just $(k!)^n$ without the $n!$?

That is exactly what we are analyzing! We proved that $\frac{(nk)!}{(k!)^n}$ is $n!$ times an integer. Because $n!$ is an integer and the number of ways to partition is an integer, the product is guaranteed to be an integer. This is the core logical leap of the problem.

Question

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

Options:

(A) $\alpha \in \mathbb{N}$ and $\beta 
otin \mathbb{N}$

(B) $\alpha 
otin \mathbb{N}$ and $\beta \in \mathbb{N}$

(C) $\alpha \in \mathbb{N}$ and $\beta \in \mathbb{N}$

(D) $\alpha 
otin \mathbb{N}$ and $\beta 
otin \mathbb{N}$

Sigma Percentile · Accepted Answer

### Analyzing the Combinatorial Structure

Imagine you are standing in a vast hall, tasked with organizing a massive collection of distinct items. You have $nk$ items, and you need to divide them into $n$ groups, with each group containing exactly $k$ items.

This is a fundamental problem in combinatorics. When you look at the expressions $\alpha = \frac{(4!)!}{(4!)^{3!}}$ and $\beta = \frac{(5!)!}{(5!)^{4!}}$, your first instinct might be panic.

You see factorials of factorials, and your brain screams that these numbers are too large to handle. But stop. In the world of JEE Advanced, whenever you see a structure like $\frac{(nk)!}{(k!)^n}$, you are not looking at a calculation problem; you are looking at a grouping problem in disguise.

### Deconstructing the Giants

Let us simplify the inner terms first. For $\alpha$, we have $4! = 24$ and $3! = 6$. Substituting these, we get:

$$\alpha = \frac{24!}{(4!)^6}$$

For $\beta$, we have $5! = 120$ and $4! = 24$. Substituting these, we get:

$$\beta = \frac{120!}{(5!)^{24}}$$

Notice the pattern? Both expressions take the form $\frac{(nk)!}{(k!)^n}$. In $\alpha$, $n=6$ and $k=4$, so $nk=24$. In $\beta$, $n=24$ and $k=5$, so $nk=120$. The structure is identical.

### The Combinatorial Insight

The number of ways to divide $nk$ distinct objects into $n$ groups of size $k$ is given by the formula:

$$W = \frac{(nk)!}{(k!)^n n!}$$

Why the $n!$ in the denominator? Because the groups are identical; swapping the groups themselves does not create a new arrangement.

Since $W$ represents a count of ways, it must be a natural number, $W \in \mathbb{N}$. Now, look at the magic of algebra. If we rearrange this formula, we get:

$$\frac{(nk)!}{(k!)^n} = W \cdot n!$$

Since $W$ is a natural number and $n!$ is a natural number, their product $W \cdot n!$ must also be a natural number.

### The Final Revelation

We have just proven that any expression of the form $\frac{(nk)!}{(k!)^n}$ is a natural number. Applying this to $\alpha$, where $n=6$ and $k=4$, we see it fits perfectly.

Applying it to $\beta$, where $n=24$ and $k=5$, it fits perfectly again. We did not need to calculate a single massive number.

We simply recognized the combinatorial soul of the expression. This is the elegance of mathematics—transforming a terrifying calculation into a simple logical conclusion. Keep this perspective, and you will find that even the most daunting problems have a hidden, beautiful simplicity.

Animated Solution for Mathematics - Permutations and Combinations: Let $α = \frac{( 4 !)!}{( 4 ! ) ^{3!}}$ and $β = \frac{( 5 !)!}{( 5 ! ) ^{4!}}$ . Then :

Select Answer:

Visualized Solution

Introduction to $α$ and $β$

Simplifying the Inner Terms of $α$

Simplifying the Inner Terms of $β$

The Combinatorial Tool: Grouping

Why the Result is an Integer

Rearranging the Formula

Applying the Logic to $α$

Applying the Logic to $β$

Final Conclusion

The Sigma Insight: Formation of Groups

Analyzing the Combinatorial Structure

Deconstructing the Giants

The Combinatorial Insight

The Final Revelation

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Animated Solution for Mathematics - Permutations and Combinations: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let α=(4!)3!(4!)!​ and β=(5!)4!(5!)!​. Then :