Prove by induction that the number of 2-subsets of an n-set A equals n(n-1)/2.
First please note that a 2-subset is a subset of a set on n elements containing exactly 2 elements, so the number of 2-subsets of an n-set is just $\displaystyle \binom{n}{2}=\frac{n(n-1)}{2}$. Now let's prove it by induction:
Base Step: When $\displaystyle n=1$, it is trivial (we have $\displaystyle \frac{1(1-1)}{2}=0$ 2-subset); when $\displaystyle n=2$, we have $\displaystyle \frac{2(2-1)}{2}=1$ 2-subset.
Inductive Step: Now lets assume when $\displaystyle n=k$, we have $\displaystyle \frac{k(k-1)}{2}$ 2-subsets. Consider the case when $\displaystyle n=k+1$ (i.e. we add one more element into the set, so how many more 2-subsets will be introduced? The answer is $\displaystyle k$ more 2-subsets are introduced. Think about this)
So the total number of 2-subsets when $\displaystyle n=k+1$ is $\displaystyle \frac{k(k-1)}{2}+k=\frac{(k+1)(k+1-1)}{2}$, this competes the inductive step.
Roy