Why does $\binom{n}{2} = \frac{n^2 - n }{2}$? In a proof in Introduction to Algorithms, the book says $\binom{n}{2} \cdot \frac{1}{n^{2}}  = \frac{n^2 - n }{2}\cdot \frac{1}{n^{2}}$, which implies $\binom{n}{2} = \frac{n^2 - n }{2}$.
Why are these equivalent? Is it a special case of a more general rule?

$\binom{n}{2}$ is the number of subsets of size $2$ from a set of size $n$. There are $n$ ways to choose the first element, and $n-1$ ways to choose the secon element. but sets are not "ordered", so we must divide by two and we have $\frac{n\cdot(n-1)}{2}$ such subsets. Thus $\binom{n}{2}=\frac{n(n-1)}{2}=\frac{n^2-n}{2}$.+

Alternatively use induction:
Base:
$\binom{2}{2}=1=\frac{2^2-2}{2}$
Induction:
$\binom{n}{2}=\binom{n-1}{2}+\binom{n-1}{n}$ by pascal recurrence.
$\binom{n-1}{2}+\binom{n-1}{2}=\frac{(n-1)^2-(n-1)}{2}+(n-1)$ by induction.
$\frac{(n-1)^2-(n-1)}{2}+(n-1)=\frac{(n-1)^2-(n-1)+2(n-1)}{2}=\frac{(n-1)^2+(n-1)}{2}=\frac{n^2-2n+1+n-1}{2}=\frac{n^2-n}{2}$ 
