J. Henry Waugh wants to know

There are six distinct rolls where all the dice show identical numbers. (1, 1, 1) through (6, 6, 6)

There are 30 distinct rolls where two numbers are identical and one is different. (6 possible pairs, from 1,1 to 6, 6, and for each pair, five numbers that don't make it a triplet.)

And there are 20 distinct rolls where all 3 of the numbers are different. (6 * 5 * 4 = 120, but each roll is counted 6 times.)

6 + 30 + 20 = 56

This can be solved by a stars and bars argument. Consider all arrangements of $(6-1) \: X$ s and $3\: I$ s such as the following: $XXIIXXXI$ These arrangements can be put into a $1$ -to- $1$ correspondence with our desired triplets. The number of $I$ s directly left to the $i$ th $X$ corresponds to the number of $i$ s in the triplet. The number of $I$ s to the right of the $5$ th $X$ corresponds to the number of $6$ s in the triplet. The above example would correspond to the triplet $(3,\: 3, \: 6)$ .

There are ${{(6-1)+3}\choose{3}}=\boxed{56}$ ways to arrange $3\: I$ s in a total of $(6-1)+3$ elements.

Direct computation:

$\text{Length}[\text{Union}[\text{Flatten}[\text{Table}[\text{Sort}[\{i,j,k\}],\{i,6\},\{j,6\},\{k,6\}],2]]] \Rightarrow 56$

English translation: generate all 216 combinations with elements sorted into ascending order (Sort) in nested lists, remove the list nesting (Flatten), remove the duplicate entries (implicit part of the Union processing) and return the length of the resultant list.