QJEEM23Ma9e4b: JEE Main 2023 Mathematics - Class 11 JEEM23Ma9e4b Solution: The number of triplets ( x , y , z ) , where x , y , z are…

We have, $x+y+z=15$

$$\begin{array}{rl}\text{ Total number of solution }& ={}^{15+3-1}{C}_{3-1}\\ \\ & ={}^{17}{C}_2=\frac{17\times 16}{1\times 2}=136\end{array}$$

Now, we need to exclude the solutions where two of $(x, y, z)$ are the same.

1) For the case $x = y \neq z$ :

$2x + z = 15$

The solutions are :

$x = 0, z = 15$

$x = 1, z = 13$

$x = 2, z = 11$

$x = 3, z = 9$

$x = 4, z = 7$

$x = 5, z = 5$ (Not valid as all are the same)

$x = 6, z = 3$

$x = 7, z = 1$

Out of these, 7 are valid.

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.

Thus, the number of triplets where all are distinct is :

$136 - 21 = 115$

There is one solution in which $\mathrm{x}=\mathrm{y}=\mathrm{z}$

Required answer $=136-21-1=114$