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?
On
$\binom{n}{2}$ 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.
$\frac{(n-1)^2-(n-1)}{2}+(n-1)=\frac{(n-1)^2-(n-1)+2(n-1)}{2}=\frac{(n-1)^2+(n-1)}{2}=\frac{n^2-2n+1+n-1}{2}=\frac{n^2-n}{2}$
Yes. In general, we have that $$ \binom{n}{k}=\frac{n!}{k!(n-k)!} $$ whenever $k\leq n$ and $0$ if $k>n$. You can see here for a more intuitive description of this using a combinatorial argument. In your case, you have $k=2$ (so you truly do have a "special case" of a more general rule). Thus, we have the following; $$\require{cancel} \binom{n}{2}=\frac{n!}{2!(n-2)!}=\frac{n(n-1)\cancel{(n-2)!}}{2!\cancel{(n-2)!}}=\frac{n(n-1)}{2}=\frac{n^2-n}{2}, $$ which is what you wanted to show. Does that clear things up?