# Q. Three numbers x, y, z are selected from the set of the first seven natural numbers such that x > 2y > 3z. How many such distinct triplets (x, y, z) are possible?

a) One triplet

b) Two triplets

c) Three triplets

d) Four triplets

Correct answer: d) Four triplets

#### Question from UPSC Prelims 2024 CSAT

**Explanation : **

# Three numbers x, y, z…

To find the number of triplets (x, y, z) from the first seven natural numbers (1 to 7) satisfying the conditions x > 2y > 3z, we’ll systematically examine possible values of z, y, and x.

## Consider z = 1:

– 3z = 3

– 2y > 3z => 2y > 3 => y > 1.5 => y >= 2

– Possible y values: 2, 3, 4, 5, 6, 7

– Compute 2y for each y and find x such that x > 2y within the set {1, 2, …, 7}:

– y = 2 => 2y = 4 => x > 4 => x = 5, 6, 7

– y = 3 => 2y = 6 => x > 6 => x = 7

– For y >= 4, 2y >= 8, but x cannot be greater than 7, so no valid x exists.

– Valid triplets for z = 1:

– (5, 2, 1)

– (6, 2, 1)

– (7, 2, 1)

– (7, 3, 1)

## Consider z >= 2:

– For z = 2:

– 3z = 6

– 2y > 6 => y > 3

– Possible y values: 4, 5, 6, 7

– 2y >= 8, but x > 2y would require x > 8, which is beyond 7.

– No valid triplets.

– Similarly, for z = 3 to 7, the required y and x values exceed 7. No valid triplets exist.

Conclusion:

Only for z = 1 do we find valid triplets, and there are four such triplets.

**Answer: Four triplets**