QUESTION

CSAT

Hard

Maths

Prelims 2019

How many triplets $(x, y, z)$ satisfy the equation $x + y + z = 6$ , where $x, y$ , and $z$ are natural numbers?

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Explanation

Step 1: Let x' = x - 1, y' = y - 1, z' = z - 1 Now, x' + y' + z' = 3

Step 2: List all possible combinations for x' + y' + z' = 3

If x' = 0: y' + z' = 3 → (0,3), (1,2), (2,1), (3,0)

If x' = 1: y' + z' = 2 → (0,2), (1,1), (2,0)

If x' = 2: y' + z' = 1 → (0,1), (1,0)

If x' = 3: y' + z' = 0 → (0,0)

Step 3: Count the total number of solutions Total = 4 (from x' = 0) + 3 (from x' = 1) + 2 (from x' = 2) + 1 (from x' = 3) = 10

The number of triplets (x, y, z) is 10. The correct answer is D. 10

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