Brute Force way of Looping Combination

How Do We LOOP C(r, n) Lets loop C(3,6)
In Combination we don't care about order so the formula of combination [n!/r!(n-r)!] gives C(3, 6) value of = 6!/3!x3! = 20

We already know we need to loop 3 times as we selected 3 items but Sometimes I confused what will the array bound and It turns out to be very simple to know the array bound

Let's take an array of [1, 2, 3, 4, 5, 6]
Here r = 3 and n = 6 = len(arr)

for the first loop the array bound will be like this
I will use go/golang and the classic for loop to write
for i := 0; i <= len(arr) - 3; i++{}
or
for i := 0; i < len(arr) - 2; i++{}
the boundary should be <= len(arr) - r or < len(arr) - (r-1)

and the second will be
for j := i+1; j <= len(arr) - 2; i++{}
or
for i := i+1; j < len(arr) - 1; i++{}
the boundary should be <= len(arr) - r-1 or < len(arr) - (r-2)

I guess now you can figure out the sequence or the formula for this.

[the second variation]