"in how many ways can we make a list of three integers"

Assuming all three integers are distinct
there are three choices for the first, 2 for the second and one for the third on the list, for a total of 3! permutations, or 6 permutations.

If two of the items are identical, then there are 3!/2!=3 ways.

If all three items are identical, then there are 3!/3!=1 way.