Relation between Triangular Numbers and Combinatorics

Question

Triangular numbers are of the form $\frac{n(n+1)}{2}$ where $n$ is a non-negative integer, or of the form $1 + 2 + 3 + .... + n$.

The no.of ways of choosing $2$ objects out of $n$ objects is also $\frac{n(n+1)}{2}$.

I was wondering whether we could count the no.of ways of choosing $2$ objects out of $n$, in the form of $1 + 2 + 3 +......+ n$.

I could not think of a way to do this.

How can we think of $n$ choose $2$ in the form of $1 + 2 + .... + n$?

Note that you get $[n(n+1)]/2$ for selection with replacement; it's actually $[n(n-1)]/2 = 1+2+\cdots+(n-1)$ without replacement. — Dan Uznanski, Aug 25 '19 at 09:48
See https://math.stackexchange.com/questions/2478616/is-there-a-combinatorial-interpretation-of-the-triangular-numbers — Jair Taylor, Aug 25 '19 at 18:07

score 1 · Accepted Answer · answered Aug 25 '19 at 09:45

1

given that the first item lexicographically that you choose is the $k$th, how many ways can the second such item be chosen?

answered Aug 25 '19 at 09:45

Dan Uznanski

Right. I can fix the 1st object, then I have (n-1) options for the second object. Then if I fix the 2nd object, I have (n-2) options as I cannot choose rhe 1st or second object and so on. Thank you. – Ram Keswani Aug 25 '19 at 09:50

score 0 · Answer 2 · answered Aug 25 '19 at 09:53

You can enumerate the objects from 1 to n. Then you can divide the possible couples into:

-Couples in which you have object n. 1

-Couples in which you have object n. 2 but not n. 1 (you have already counted couple $\{ 1,2\}$ )

-Couples in which you have object n. 3 but not n. 1 and n. 2

And so on...

If you sum cardinalities of this groups you find $1+2+3+4+5...+n$

2 Answers2