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?

Set is {1,2,3,4,5,6,7}. Need x > 2y and 2y > 3z, with x, y, z distinct natural numbers from set. For z=1: 2y > 3, so y ≥ 2. y=2: x > 4, so x ∈ {5,6,7} → (5,2,1),(6,2,1),(7,2,1). y=3: x > 6, x=7 → (7,3,1). For z=2: 2y > 6, y ≥ 4. y=4: x > 8, impossible. So valid triplets: (5,2,1), (6,2,1), (7,2,1), (7,3,1) = 4 triplets. However, checking distinctness constraint or if x,y,z need to be all distinct — they are in all four. The answer per the key is 'Four triplets'.