We need the number of ordered triplets $(x,y,z)$ such that:

• $x,y,z$ are distinct non-negative integers
• $x+y+z=15$

First, count all ordered non-negative integer solutions of $x+y+z=15.$ Using stars and bars, the number of solutions is $\binom{15+3-1}{3-1}=\binom{17}{2}=136.$

Now subtract the solutions where at least two variables are equal.

Since $x,y,z$ are required to be distinct, we must remove all cases with:

• $x=y$
• $y=z$
• $z=x$

Count solutions with $x=y$. Let $x=y=a$. Then $2a+z=15.$ Since $z\ge 0$, we need $a=0,1,2,3,4,5,6,7$. This gives $8$ ordered solutions.

Similarly:

• $y=z$ gives $8$ solutions
• $z=x$ gives $8$ solutions

So total with at least one specified equality is $8+8+8=24.$

Check overlap: could all three be equal? If $x=y=z$, then $3x=15 \Rightarrow x=5,$ so $(5,5,5)$ is one solution.

This solution has been counted in all three equalities, so by inclusion-exclusion: $24-3+1?$

More carefully, $|x=y \cup y=z \cup z=x| = 8+8+8 -1-1-1 +1 = 22,$ because the only pairwise intersection is $(5,5,5)$, and the triple intersection is also $(5,5,5)$.

Therefore, the number of ordered triplets with all distinct entries is $136-22=114.$

Hence the correct option is $\boxed{114}.$

Comparing with the stored correct answer:

• Stored answer: D
• Derived answer: D So they agree.