Q49. 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 numbers x, y, z must be distinct natural numbers from the set {1, 2, 3, 4, 5, 6, 7}. The given condition is x > 2y > 3z. Let's start by finding possible values for z, as it has the tightest constraint (3z).

Case 1: z = 1 3z = 3. So, 2y > 3, which means y > 1.5. Possible values for y are 2, 3. - If y = 2: 2y = 4. So, x > 4. Possible values for x are 5, 6, 7 (since x must be distinct from y and z, and from the set).

Triplets: (5, 2, 1), (6, 2, 1), (7, 2, 1) - (3 triplets) - If y = 3: 2y = 6. So, x > 6. Possible value for x is 7 (since x must be distinct from y and z, and from the set). Triplet: (7, 3, 1) - (1 triplet) Case 2: z = 2 3z = 6.

So, 2y > 6, which means y > 3. Possible values for y are 4, 5, 6, 7. - If y = 4: 2y = 8. So, x > 8. No possible value for x from the set {1, ..., 7}. Since no x is possible for y=4, no further values of y (5, 6, 7) will yield a solution either.

Therefore, the distinct triplets are (5, 2, 1), (6, 2, 1), (7, 2, 1), and (7, 3, 1). There are a total of 4 such distinct triplets. This question tests logical reasoning and number properties.