The number of triplets x ,   y ,   z where x ,   y ,   z are distinct non negative…

Given,

$x+y+z=15$

Now we know that,

Non-negative integral solution of equation $a+b+c=n$ is given by ${}^{n+3-1}C_{3-1}$ where $a=b=c \& a=b\ne c$ are also possibilities

So by above formula we get,

Total number of non-negative solution will be,  ${}^{15+3-1}C_{3-1}={}^{17}C_2$

Now solving If any of these $2$ are equal
So, the equation will become$x+2y=15$

Now finding possible cases we get,

$\begin{array}{ll}y=0& x=15\end{array}$

$\begin{array}{ll}y=1& x=13\end{array}$

$\begin{array}{ll}y=2& x=11\end{array}$

$\begin{array}{ll}y=3& x=9\end{array}$

$\begin{array}{ll}y=4& x=7\end{array}$

$y=5 x=5\to x=y=z=5$

$y=6 x=3$

$y=7 x=1$

So, total possibilities where $x, y \& z$ are distinct will be,

$={}^{17}C_2-{}^{3}C_2\times 8+2$

{Note adding $2$ because the cases $x=y=z$ is subtracted three times}

$=136-24+2=114$