Induction Proof

Not sure why you want to use induction to do this but here you go in a fast and loose manner:

Assume it's true for a set of size n. Examine a set of size n+1. Notice this set of size n+1 is just an n-set with a new element added to it. Call that new element 'A'. Now what does a 2-subset of the n+1-set look like? Either it has A or it doesn't. Thus, using our inductive hypothesis, we see there are:

[tex]\frac{n*(n-1)}{2} + n = \frac{(n+1)*n}{2}[/tex]

2-subsets. And that's exactly what we want to see.

 

you mean, considering a set of n elements the number of subsets with 2 elements obtained from that set is equal to n(n-1)/2?

 

if so, it is easy to see that the first combine with n-1 elements, the second with n-2 elements, and so on

 

Yes That gives a direct formula but what the poster needed was a step for an inductive proof which is what Rodigee gave since the new element (n+1) combines with the first n elements to form n more sets of two elements.