Today when I was reading my algebra book I encountered this following problems, given the set $$A = \{1, 2, \dots, n-1, n \}$$
I realized that if I just list all the possible $(i,j)$ such that $i,j \in A$ and $i < j$, then this gives you exactly all the possible ways of pick $2$ elements from $A$ without repetition.
This is intuitively correct, but is there a way to proves this? I have zero experience with combinatorial math, but here is my attempt:
Given $(i,j)$ such that $i,j \in A$ and $i < j$, this is clearly a way of pick $2$ elements from $A$. On the other hand, if I choose $i$ and $j$ from $A$, by possible rearrangement, we can assume that $i < j$, then this gives you an element $(i,j)$ such that $i,j \in A$ and $i < j$. This gives you a bijective correspondence between all $(i,j)$ where $i,j \in A$ and $i < j$, and all possible way of pick $2$ elements from $A$.
is this proof correct?
Further, can I generalize this to the following statement, the list of all $(i,j,k)$ where $i,j,k \in A$ and $i<j<k$ would gives you all possible ways of pick $3$ elements from $A$?