The number of triplets $$(x, y, z)$$ where $$x, y, z$$ are distinct non negative integers satisfying $$x + y + z = 15$$, is

We need to find the number of triplets $$(x, y, z)$$ of distinct non-negative integers satisfying $$x + y + z = 15$$.

Total solutions without distinctness constraint.

The number of non-negative integer solutions of $$x + y + z = 15$$ is:

$$\binom{15 + 2}{2} = \binom{17}{2} = 136$$

Subtract solutions where at least two variables are equal.

Let A = {x = y}, B = {y = z}, C = {x = z}.

$$|A|$$: If $$x = y$$, then $$2x + z = 15$$, so $$x$$ ranges from 0 to 7. That gives 8 solutions. Similarly $$|B| = |C| = 8$$.

$$|A \cap B|$$: If $$x = y = z$$, then $$3x = 15$$, so $$x = 5$$. That gives 1 solution. Similarly for all pairwise intersections and the triple intersection: $$|A \cap B| = |A \cap C| = |B \cap C| = |A \cap B \cap C| = 1$$.

By inclusion-exclusion:

$$|A \cup B \cup C| = 8 + 8 + 8 - 1 - 1 - 1 + 1 = 22$$

Distinct solutions.

$$136 - 22 = 114$$

The number of triplets is 114.