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?

The correct option is Four triplets.

Let's find the number of distinct triplets $(x,y,z)$ such that $x,y,z$ are selected from the first seven natural numbers, which means:
$x,y,z\in \{1,2,3,4,5,6,7\}$
and they satisfy the inequality:
$x>2y>3z$

We can analyze the possible values step-by-step starting from the smallest variable, $z$:

Case 1: Let $z=1$
If $z=1$, then the inequality becomes:
$x>2y>3\left(1\right)\Rightarrow x>2y>3$
This implies that:
$2y>3\Rightarrow y>1.5$
Since $y$ must be a natural number from the set, the possible values for $y$ can be 2, 3, etc. Let's analyze each possible value of $y$:

Sub-case 1a: Let $y=2$
Then $2y=4$. The inequality $x>2y$ becomes:
$x>4$
Since $x$ is a natural number less than or equal to 7, the possible values for $x$ are:
$x\in \{5,6,7\}$
This gives us 3 distinct triplets:
1. $(5,2,1)$
2. $(6,2,1)$
3. $(7,2,1)$

Sub-case 1b: Let $y=3$
Then $2y=6$. The inequality $x>2y$ becomes:
$x>6$
Since $x$ is a natural number less than or equal to 7, the only possible value for $x$ is:
$x=7$
This gives us 1 distinct triplet:
4. $(7,3,1)$

Sub-case 1c: Let $y\ge 4$
Then $2y\ge 8$. The inequality $x>2y$ would require $x>8$, which is impossible since the maximum value in our set is 7.

Case 2: Let $z\ge 2$
If $z\ge 2$, then:
$3z\ge 6$
The inequality becomes:
$x>2y>3z\ge 6\Rightarrow x>2y>6$
This means:
$2y>6\Rightarrow y>3$
If $y\ge 4$, then $2y\ge 8$, which requires $x>8$. This is not possible as $x\le 7$. Thus, no valid triplets exist for $z\ge 2$.

Combining all possible cases, we have exactly 4 valid triplets:
$(5,2,1),(6,2,1),(7,2,1),\text{and}(7,3,1)$.