JEE Mathematics 2023 Question with Solution

Given: $x + y + z = 15$, where $x, y, z$ are distinct non-negative integers.

Find: The number of triplets $(x, y, z)$ satisfying the condition.

First count all non-negative integer solutions:

$$\text{Total number of solutions} = \binom{15 + 3 - 1}{3 - 1} = \binom{17}{2} = 136$$

This counts all solutions, including cases where some variables are equal.

Now consider the case $x = y = z$:

$$3x = 15 \Rightarrow x = 5$$

So there is exactly $1$ such solution.

Next consider the cases where two variables are equal. Suppose $x = y$. Then

$$2x + z = 15$$

so

$$z = 15 - 2x$$

The distinct solutions occur when $x$ takes values from $1$ to $7$. Hence there are $7$ such solutions.

Therefore, the required number of distinct solutions is

$$136 - 1 - 7 = 114$$

Therefore, the number of distinct non-negative integer triplets is $114$. The correct option is B.

Given: $x + y + z = 15$ with $x, y, z$ distinct non-negative integers.

Find: The number of valid ordered triplets.

The solution uses stars and bars to count all non-negative integer solutions first:

$$\binom{17}{2} = 136$$

Then subtract solutions that are not distinct.

1. All three equal:

$$x = y = z$$ $$3x = 15$$ $$x = 5$$

So one invalid triplet occurs.

2. Two equal: Assume

$$x = y$$

Then

$$2x + z = 15, \quad z = 15 - 2x$$

From the extracted solution, $x$ can take values from $1$ to $7$, giving $7$ such cases.

Subtracting these from the total:

$$136 - 1 - 7 = 114$$

Hence the required number of triplets is $114$.