The number of triplets (x, y, z),where x, y, | vedclass

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(B) The total number of non-negative integer solutions to $x+y+z=15$ is given by the stars and bars formula: $\binom{15+3-1}{3-1} = \binom{17}{2} = \f

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$114$

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(B) The total number of non-negative integer solutions to $x+y+z=15$ is given by the stars and bars formula: $\binom{15+3-1}{3-1} = \binom{17}{2} = \frac{17 	imes 16}{2} = 136$.Let $S$ be the set of all non-negative integer solutions. We want to find the number of solutions where $x, y, z$ are distinct.First,find the number of solutions where at least two variables are equal:Case $1$: $x=y$. Then $2x+z=15$. Since $z \ge 0$,$2x \le 15$,so $x \in \{0, 1, 2, 3, 4, 5, 6, 7\}$. This gives $8$ solutions.By symmetry,the cases $y=z$ and $x=z$ also yield $8$ solutions each.Total solutions with at least two variables equal is $8+8+8 = 24$.Case $2$: $x=y=z$. Then $3x=15$,so $x=5$. This gives $1$ solution $(5, 5, 5)$.Note that the solution $(5, 5, 5)$ is counted in all three sub-cases of Case $1$ ($x=y$,$y=z$,$x=z$).Using the Principle of Inclusion-Exclusion,the number of solutions where at least two are equal is $8+8+8 - 2(1) = 22$.Number of solutions where $x, y, z$ are distinct = (Total solutions) - (Solutions where at least two are equal) = $136 - 22 = 114$.

The number of triplets $(x, y, z)$,where $x, y, z$ are distinct non-negative integers satisfying $x+y+z=15$,is

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