Prove by induction that the number of 2-subsets of an n-setAequalsn(n-1)/2.

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- Mar 14th 2008, 02:47 PMst4rl1ightInduction Proof.........
Prove by induction that the number of 2-subsets of an n-set

**A**equals**n(n-1)/2**. - Mar 14th 2008, 08:48 PMroy_zhang
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