Let $G$ be a finite group with $n$ elements with initial order $g_1,g_2,...,g_n$ and let's create the group multiplication table of $G$ with this initial order.
In that table you will have $n$ rows and every element appear exactly one time in one row.
Let $s_i$ be the product of elements of the $i$th row in the order of rows.
The question is how many different $s_i$ can we obtain ?
My motivation depends on this question and its answer is $|G'|$, i.e., the number of elements of the commutator group. Thus, in our case this number can be at most $|G'|$.
I also have doubt that whether this number depends on initial order of the elements of $G$ or not.
Thanks.
Edit: For $G=S_3$ answer is $3=|G'|$ which is the upper bound.
I guessed as you did that this does depend on the ordering of the $g_i$ and ran a program to check. I found (if I have not made a error) that the number of $s_i$ may differ. In the dihedral group $D_4$, with the following ordering
()
(2,4)
(1,2)(3,4)
(1,2,3,4)
(1,3)
(1,3)(2,4)
(1,4,3,2)
(1,4)(2,3)
switching the last two elements changes the number of $s_i$ from 1 to 2. That being said, I would guess that there is enough "wiggle room" that one cannot say much more than the number of distinct $s_i$ lies somewhere in 1 and $|G'|$, though it would be relatively interesting if this were not the case.