Explanation:

- The set of the first seven natural numbers is {1, 2, 3, 4, 5, 6, 7}.

- We need to find triplets (x, y, z) such that \(x > 2y > 3z\).

- Let's try different values for z.

- If z = 1:

- \(3z = 3\), so \(2y > 3\). Possible y values are 2, 3.

- If y = 2, \(2y = 4\); thus x > 4, possible x values are 5, 6, 7 (Triplets: (5, 2, 1), (6, 2, 1), (7, 2, 1)).

- If y = 3, \(2y = 6\); thus x > 6, possible x value is 7 (Triplet: (7, 3, 1)).

- If z = 2:

- \(3z = 6\), so \(2y > 6\). The smallest y could be is 4.

- If y = 4, \(2y = 8\); thus x > 8, but there’s no x in the set satisfying this.

- No valid triplet for z = 2 or higher given y's constraints for these z's.

- Conclusion: There are a total of 4 valid triplets: (5, 2, 1), (6, 2, 1), (7, 2, 1), and (7, 3, 1).

- Correct Answer: Option: 4 - Four triplet

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