Prove by permutation or otherwise (n^2)! / (n!)^n is an integer (n ∈ I+).

Q: Wait, why can't we just use induction? It seems like a standard proof problem.

Induction is a powerful tool, but it often hides the 'why'. In JEE, we want to see the combinatorial beauty. Induction would involve complex factorial manipulation, whereas this combinatorial proof is elegant and intuitive.

Q: Does the order of the groups matter?

Great question! In the multiset formula, we are counting distinct arrangements of the *items*. Since the items within each group are identical, swapping two items of the same color doesn't create a new arrangement. That's why we divide by $(n!)^n$.

Q: Is this proof valid for all $n$?

Yes, $n$ must be a positive integer ($n \in I^+$). If $n=1$, the expression is $1!/1! = 1$, which is an integer. The logic holds for all positive integers.

Question

Prove by permutation or otherwise $\frac{(n^2)!}{(n!)^n}$ is an integer ($n \in I^+$).

Sigma Percentile · Accepted Answer

### The Beauty of Combinatorial Logic

Welcome, future engineers! Today, we are going to dismantle a problem that looks like a nightmare but is actually a beautiful piece of logic. We are looking at the expression:

$$\frac{(n^2)!}{(n!)^n}$$

At first glance, it is a mess of factorials. But in the world of JEE Advanced, we don't just calculate; we interpret. Let's embark on this journey.

### Phase 1: Deconstructing the Expression

When you see a factorial expression like $\frac{(n^2)!}{(n!)^n}$, your first instinct might be to reach for algebraic identities or induction. While those paths exist, they are often tedious.

Instead, let's look at the structure. The numerator is $(n^2)!$, which is the factorial of a total number of objects, $N = n^2$. The denominator is $(n!)^n$, which is $n!$ multiplied by itself $n$ times.

This structure screams **multiset permutation**.

### Phase 2: The Multiset Permutation Theorem

Recall the fundamental theorem of permutations for multisets. If you have $N$ objects, where $p$ are of one type, $q$ are of another, and so on, the number of distinct arrangements is given by:

$$\frac{N!}{p! q! r! \dots}$$

This formula is the key to our proof. Because this formula counts the number of ways to arrange objects, the result **must be an integer**. You cannot have a fraction of an arrangement!

### Phase 3: Building the Physical Model

Imagine you are standing in front of a table. You have $n$ distinct groups of balls. For each group, you have exactly $n$ identical balls.

For example, if $n=3$, you have 3 blue balls, 3 green balls, and 3 red balls. The total number of balls is $N = n + n + \dots + n$ ($n$ times), which equals $n^2$.

Now, if you want to arrange all these $n^2$ balls in a single row, how many distinct ways can you do it?

### Phase 4: The "Aha!" Moment

Using our multiset formula, the total number of arrangements is the factorial of the total number of balls, divided by the factorial of the count of each identical group.

So, we have:

$$\frac{(n^2)!}{n! \cdot n! \cdot \dots \cdot n!}$$

Since there are $n$ groups, we divide by $n!$ exactly $n$ times. This gives us the expression:

$$\frac{(n^2)!}{(n!)^n}$$

### Conclusion: The Power of Perspective

Because this expression represents the number of distinct ways to arrange these $n^2$ balls, it must be a whole number. There is no other physical possibility.

By shifting our perspective from abstract algebra to physical counting, we have proven that $\frac{(n^2)!}{(n!)^n}$ is an integer for all $n \in I^+$.

This is the essence of JEE Advanced: finding the simplest, most elegant path through the complexity. Keep practicing, and keep looking for the story behind the math!

Animated Solution for Mathematics - Permutations and Combinations: Prove by permutation or otherwise $\frac{( n ^{2} )!}{( n ! ) ^{n}}$ is an integer ( $n \in I^{+}$ ).

Visualized Solution

The Objective: Prove $\frac{( n ^{2} )!}{( n ! ) ^{n}} \in I$

Permutations of a Multiset

Counting Must Be an Integer

Defining the Objects

Adding More Groups

$n$ Distinct Groups

Total Number of Objects

Simplifying Total Objects

Arranging the Objects

Accounting for Identical Items

The Final Expression

The Conclusion

The Sigma Insight: Linear Permutations

The Beauty of Combinatorial Logic

Phase 1

Phase 2

Phase 3

Phase 4

Conclusion

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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; } Prove by permutation or otherwise (n!)n(n2)!​ is an integer (n∈I+).