How many combinations of $3$-digit numbers are there from $000-999$, where the ordering of the digits don't matter. In this case, $112$ would be the same as $121$ and $211$ and so the set would contain only one of them.
I think I might have worked it out by partitioning it into $3$ cases:
case 1: all digits are the same $= 10$
case 2: $1$ digit is the same, $2$ digits are different $= 10 \cdot 9 = 90$
case 3: all digits are different $= \binom{10}{3} = 120$
so case 1 + case 2 + case 3 $= 220$