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So:

$$\binom{n}{2} = \frac{n!}{2!(n-2)!}$$

Using Stirling's approximation we have:

$$\frac{\sqrt{2 \pi n}(\frac{n}{e})^n}{[\sqrt{2 \pi 2}(\frac{2}{e})^2][\sqrt{2 \pi (n-2)}(\frac{(n-2)}{e})^{(n-2)}]}$$

Removing constants and simplifying:

$$\frac{(\frac{n}{e})^n}{(\frac{(n-2)}{e})^{(n-2)}} = \frac{n^n}{e^n} \times \frac{e^{n-2}}{(n-2)^{n-2}} = \frac{n^n}{(n-2)^{(n-2)}}$$

Not sure how to proceed.

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  • 10
    $\begingroup$ $$\binom{n}2=\frac{n(n-1)}2$$ $\endgroup$ Jun 22 '15 at 21:21
  • $\begingroup$ Can you show me the derivation? $\endgroup$
    – Chris
    Jun 22 '15 at 21:22
  • 10
    $\begingroup$ Stirling's approximation to compute $n(n-1)$? Are you joking? $\endgroup$ Jun 22 '15 at 21:23
  • 1
    $\begingroup$ Funnily enough you nearly had it with your (rather convoluted) method: $$\frac{n^n}{(n-2)^{n-2}} \sim \frac{n^n}{n^{n-2}} = n^2$$ $\endgroup$ Jun 22 '15 at 21:42
  • 1
    $\begingroup$ Well $$\frac{n^n}{(n-2)^{n-2}} = \frac{n^n}{n^{n-2}} \cdot \left( \frac{n}{n-2} \right)^{n-2}$$ And $\left( \frac{n}{n-2} \right)^{n-2} \to e^2$ as $n \to \infty$, which is constant. (I probably abused the $\sim$ symbol, but you catch my drift.) $\endgroup$ Jun 22 '15 at 22:00
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As pointed out in the comments a simplification obviates the need for Sterling's approximation:

$$\binom{n}{2} = \frac{n!}{2!(n-2)!} = \frac{n(n-1)}{2} = \frac{n^2}{2} - \frac{n}{2} = \mathcal{O}(n^2)$$

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