The number of triplets (x, y, z) where x, y, z are distinct non...

Given,

$x + y + z = 15$

Now we know that,

Non-negative integral solution of equation $a + b + c = n$ is given by $C3 - 1 n + 3 - 1$ where $a = b = c \text{ and } a = b \ne c$ are also possibilities

So by above formula we get,

Total number of non-negative solution will be, $C3 - 1 15 + 3 - 1 = C2 17$

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

Now finding possible cases we get,

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

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

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

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

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

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

$y = 6x = 3$

$y = 7x = 1$

So, total possibilities where $x, y \text{ and } z$ are distinct will be,

$= C2 17 - C2 3 \times 8 + 2$

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

$= 136 - 24 + 2 = 114$