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This problem , I assume can be proved using induction, however I am trying to find another way.

Is there a simple combinatorial approach? One notices that $(n!)^2$ is equal to the number of permutations of size n squared, and that $n^n$ is the number of redundant combinations where there are n spaces and n choices.

Any help would be much appreciated.

Thanks

Hummus
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2 Answers2

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This is not combinatorial, but note that $$(n!)^2=\prod_{k=1}^n k(n+1-k).$$ (We are in essence using the "Baby Gauss" trick.) But if $k$ is not $1$ or $n$, we have $k(n+1-k)\gt n$.

André Nicolas
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  • Thanks for your help- could you please explain how you achieved the formula? I'm slightly confused? – Hummus Feb 16 '14 at 18:24
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    We have $(n!)^2=(1\cdot 2\cdot 3\cdots n)(n\cdot (n-1)\cdots 2\cdot 1)$ (I have reversed the second factorial). This is $(1\cdot n) (2\cdot (n-1))\cdots (n\cdot 1)$. Just like finding $1+2+\cdots +n$ by reversing and adding term by term, the usual trick for finding $1+2+\cdots +n$, except in our case we get inequalities. – André Nicolas Feb 16 '14 at 18:35
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divide $(n!)^2 > n^n$ by $n!$ to get

$$n! = 1 \times 2 \times \ldots \times (n-1) \times n > \frac{n}{n} \times \frac{n}{n-1} \times \ldots \times \frac{n}{2} \times \frac{n}{1}$$

It is a bit of simple algebraic manipulation to show that each term on the lhs is greater or equal to the corresponding term on the rhs, or $\frac{n}{n-k} < k+1 $ for $k \in \overline{0, n-1}$

huh
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