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

• The condition is x > 2y > 3z, which implies x > 2y and 2y > 3z.

• Since z is a natural number, the minimum value for z is 1. If z = 1, then 3z = 3, so 2y must be greater than 3. The smallest integer y satisfying 2y > 3 is y = 2 (since 2*2 = 4 > 3).

• With y = 2, 2y = 4. The condition x > 4 requires x to be 5, 6, or 7. This gives three triplets: (5, 2, 1), (6, 2, 1), and (7, 2, 1).

• If z = 2, then 3z = 6, so 2y must be greater than 6. The smallest integer y satisfying 2y > 6 is y = 4 (since 2*4 = 8 > 6).

• With y = 4, 2y = 8. The condition x > 8 requires x to be greater than 8. However, the maximum number in the set is 7, so no such x exists for z = 2 or higher.

• Therefore, there are exactly four distinct triplets: (5, 2, 1), (6, 2, 1), (7, 2, 1), and (8, 4, 2) is invalid, wait, let me re-evaluate y=3. If y=3, 2y=6. 6 > 3z implies 3z < 6, so z can be 1. If z=1, 3z=3, 2y=6 > 3 is true. Then x > 6, so x can be 7. Triplet (7, 3, 1). Let's list all systematically.

• Case z=1: 3z=3. 2y > 3 => y >= 2.

• If y=2, 2y=4. x > 4 => x in {5, 6, 7}. Triplets: (5,2,1), (6,2,1), (7,2,1).

• If y=3, 2y=6. x > 6 => x in {7}. Triplet: (7,3,1).

• If y=4, 2y=8. x > 8 => No x in set.

• Case z=2: 3z=6. 2y > 6 => y >= 4.

• If y=4, 2y=8. x > 8 => No x in set.

• Total triplets are (5,2,1), (6,2,1), (7,2,1), (7,3,1). There are 4 triplets.