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

We have, $x+y+z=15$ <br/><br/>$$ \begin{aligned} \text { Total number of solution } & ={ }^{15+3-1} C_{3-1} \\\\ & ={ }^{17} C_2=\frac{17 \times 16}{1 \times 2}=136 \end{aligned} $$ <br/><br/>Now, we need to exclude the solutions where two of $(x, y, z)$ are the same. <br/><br/>1) For the case $x = y \neq z$ : <br/><br/>$ 2x + z = 15 $ <br/><br/>The solutions are : <br/><br/>$ x = 0, z = 15 $ <br/><br/>$ x = 1, z = 13 $ <br/><br/>$ x = 2, z = 11 $ <br/><br/>$ x = 3, z = 9 $ <br/><br/>$ x = 4, z = 7 $ <br/><br/>$ x = 5, z = 5 $ (Not valid as all are the same) <br/><br/>$ x = 6, z = 3 $ <br/><br/>$ x = 7, z = 1 $ <br/><br/>Out of these, 7 are valid. <br/><br/>Similarly, for the cases $y = z \neq x$ and $z = x \neq y$, there will be 7 valid solutions for each, so a total of $ 7 \times 3 = 21 $ solutions where two of the variables are equal. <br/><br/>Thus, the number of triplets where all are distinct is : <br/><br/>$ 136 - 21 = 115 $ <br/><br/>There is one solution in which $\mathrm{x}=\mathrm{y}=\mathrm{z}$ <br/><br/>Required answer $=136-21-1=114$