The number of triplets (x, y, z) where x, y, z are distinct non negative integers satisfying x + y +

Q: Wait, why do we subtract the case where $x=5$ in the 'exactly two equal' scenario?

That is a fantastic observation! When we solve $2x + z = 15$, we find all pairs $(x, z)$ such that $x=y$. This includes the case where $x=5$, which forces $z=5$. This specific triplet $(5, 5, 5)$ has all three variables equal. Since we are counting 'exactly two equal' cases separately, we must remove this triplet to avoid double-counting it with the 'all three equal' case.

Q: Why can't we just use permutations to solve this?

Permutations are great when we are arranging distinct items, but here we are dealing with integer partitions where the order of variables $(x, y, z)$ matters in the equation $x+y+z=15$. The 'Stars and Bars' method naturally accounts for the order of variables. If you tried to use permutations, you would have to account for the symmetry of the variables, which is much more complex than the complementary counting approach we used.

Q: Is there a direct way to solve this without complementary counting?

Yes, you could manually list the partitions of 15 into 3 distinct non-negative integers (e.g., $0+1+14, 0+2+13, \dots$) and then permute them. However, that is extremely prone to error and time-consuming. Complementary counting is the 'JEE way'—it is elegant, systematic, and significantly reduces the chance of missing a case.

Question

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

Options:

(A) 80

(B) 136

(C) 114

(D) 92

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The problem asks us to find the number of non-negative integer solutions to the equation:

$$x + y + z = 15$$

subject to the constraint that $x, y, z$ must be **distinct**. We will approach this by first calculating the total number of non-negative solutions and then subtracting the cases where the variables are not distinct.

### The Total Universe

Ignoring the distinct constraint, we use the **Stars and Bars** method. For an equation $x_1 + x_2 + \dots + x_r = n$, the number of non-negative integer solutions is given by the formula:

$$\binom{n+r-1}{r-1}$$

Here, $n = 15$ and $r = 3$. Substituting these values, we get:

$$\binom{15+3-1}{3-1} = \binom{17}{2}$$

Calculating this value:

$$\binom{17}{2} = \frac{17 	imes 16}{2 	imes 1} = 136$$

There are 136 total non-negative integer solutions.

### The Anatomy of the 'Not-Distinct' Trap

To find the number of distinct triplets, we use **complementary counting**. We must subtract the cases where the variables are not distinct.

These unwanted cases fall into two categories:
1. All three variables are equal ($x = y = z$).
2. Exactly two variables are equal (e.g., $x = y 
eq z$).

### Analyzing the Unwanted Cases

First, consider the case where all three are equal: $x = y = z$. Substituting into the original equation:

$$3x = 15 \implies x = 5$$

This yields exactly one triplet: $(5, 5, 5)$.

Next, consider the case where exactly two variables are equal. Let $x = y 
eq z$. The equation becomes:

$$2x + z = 15$$

Since $z \ge 0$, we have $2x \le 15$, meaning $x$ can be any integer from $0$ to $7$. This provides 8 possible pairs for $(x, z)$.

However, if $x = 5$, then $z = 5$, which results in $(5, 5, 5)$. Since we already counted this in the "all three equal" category, we exclude it. This leaves $8 - 1 = 7$ valid cases where exactly two variables are equal.

Because the equal pair could be $(x, y)$, $(y, z)$, or $(z, x)$, there are 3 such scenarios. Thus, the total number of cases where exactly two variables are equal is:

$$3 	imes 7 = 21$$

### Final Calculation

The total number of unwanted cases is the sum of the "all three equal" case and the "exactly two equal" cases:

$$1 + 21 = 22$$

Finally, we subtract the unwanted cases from the total universe to find the number of distinct triplets:

$$136 - 22 = 114$$

The total number of distinct non-negative integer triplets is **114**.

Animated Solution for Mathematics - Permutations and Combinations: The number of triplets $(x, y, z)$ where $x, y, z$ are distinct non negative integers satisfying $x + y + z = 15$ , is

Select Answer:

Visualized Solution

Understanding the Problem

The Strategy

Total Solutions Formula

Calculating Total Solutions

Computing $^{17} C_{2}$

Breaking Down Unwanted Cases

Case 1: All Three Equal

Counting Case 1

Case 2: Exactly Two Equal

Solving $2 x + z = 15$

Removing the Overlap

Permuting the Pairs

Total Unwanted Cases

Final Calculation

The Sigma Insight: Combinations and Selection

Analyzing the Setup

The Total Universe

The Anatomy of the 'Not-Distinct' Trap

Analyzing the Unwanted Cases

Final Calculation

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Animated Solution for Mathematics - Permutations and Combinations: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 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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