It's ordered triplets, not unordered triplets

This is a stars and bars problem. Referring to Theorem 1 in the link, with $n = 10$ and $k = 3$ we find that the solution is

$\dbinom{10 - 1}{3 - 1} = \dbinom{9}{2} = \boxed{36}.$

Consider each of the triplets that add up to 10.

1. (1,1,8)*
2. (1,2,7)
3. (1,3,6)
4. (1,4,5)
5. (2,2,6)*
6. (2,3,5)
7. (2,4,4)*
8. (3,3,4)*

Here a rearrangement of numbers in any of these sets also gives us the sum of 10.

For the triplets with * marked, each case will have $\boxed{3 }$ combinations possible. These sets have two entries being the same number. Only the distinct number needs to be placed in any one out of three, thus leading to 3 different solutions.

While for the remaining triplets the entries are distinct , and $\boxed{6 }$ combinations are possible for each of them. Here two numbers are to be filled in two out of three places.

In total we have $4*3+4*6= \boxed{ 36}$ solutions.

If we dispose ten 1 in a line, the number of spaces between each one is 9. Then we only have to choose two of those nine spaces.

An example:

1 1 | 1 1 1 1 | 1 1 1 1 ---> the solution is (2, 4, 4)

1 1 1 | 1 1 1 1 1 | 1 1 ---> the solution is (3, 5, 2)

...

Thus, a solution is 9C2 = 36.