We are given the set of the first seven natural numbers: ${1, 2, 3, 4, 5, 6, 7}$. We need to find the number of distinct triplets $(x, y, z)$ such that:

• $x > 2y$
• $2y > 3z$

Let's analyze each possible value of $z$ and find corresponding values of $y$ and $x$.

Case 1: $z = 1$

• For $2y > 3z \rightarrow 2y > 3 \rightarrow y > 1.5$, so $y \geq 2$.
• For $x > 2y \rightarrow x > 2y$, so possible values of $x$ are greater than $2y$.

Checking for possible values:

• $y = 2 \rightarrow x > 4$ ($x = 5, 6, 7$)
• $y = 3 \rightarrow x > 6$ ($x = 7$)
• Total triplets: $(5, 2, 1)$, $(6, 2, 1)$, $(7, 2, 1)$, $(7, 3, 1)$

Case 2: $z = 2$

• For $2y > 3z \rightarrow 2y > 6 \rightarrow y > 3$, so $y \geq 4$.
• For $x > 2y \rightarrow x > 2y$, so possible values of $x$ are greater than $2y$.

Checking for possible values:

• $y = 4 \rightarrow x > 8$ (no possible $x$)
• Total triplets: None for $z = 2$.

Case 3: $z = 3$

• For $2y > 3z \rightarrow 2y > 9 \rightarrow y > 4.5$, so $y \geq 5$.
• For $x > 2y \rightarrow x > 2y$, so possible values of $x$ are greater than $2y$.

Checking for possible values:

• $y = 5 \rightarrow x > 10$ (no possible $x$)
• Total triplets: None for $z = 3$.

Conclusion: From the calculations above, the possible triplets are:

• $(5, 2, 1)$
• $(6, 2, 1)$
• $(7, 2, 1)$
• $(7, 3, 1)$

Thus, there are 4 distinct triplets.

Answer: D. Four triplets