Prove by permutation or otherwise...

MathematicsLinear PermutationsJEE Advanced 2004Moderate

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

Visualized Solution (Hindi)

The Objective: Prove .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } \frac{( n ^{2} )!}{( n ! ) ^{n}} \in I

Goal: Prove $\frac{( n ^{2} )!}{( n ! ) ^{n}}$ is an integer for $n \in I^{+}$ .
Approach: Use the combinatorial interpretation of permutations of a multiset.

Permutations of Multisets

Tool: Formula for permutations of $N$ objects with groups of identical items.
$Ways = \frac{N !}{p ! q ! r ! \dots}$
Since the number of ways to arrange objects must be a natural number, this ratio is always an integer.

Defining the Objects

Consider $n$ distinct sets of objects.
Each set contains exactly $n$ identical objects.
Total objects $N = n + n + \dots + n (n times) = n^{2}$ .

Applying the Formula

Total objects $N = n^{2}$ .
Identical groups: $n$ groups, each with $n$ identical items.
Number of arrangements $= \frac{( n ^{2} )!}{n ! \cdot n ! \cdot \dots \cdot n ! (n times)}$
This simplifies to $\frac{( n ^{2} )!}{( n ! ) ^{n}}$ .

The Conclusion

The expression $\frac{( n ^{2} )!}{( n ! ) ^{n}}$ represents the number of ways to arrange $n^{2}$ objects.
By definition, the number of arrangements must be a whole number.
Hence, $\frac{( n ^{2} )!}{( n ! ) ^{n}}$ is an integer for all $n \in I^{+}$ .
Key Takeaway: Combinatorial interpretations can turn complex fractions into simple counting problems.

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