Q. 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?

Explanation:

Distinct triplets (x, y, z)

Let’s approach this step-by-step:

1) First, we need to identify the set of the first seven natural numbers:
{1, 2, 3, 4, 5, 6, 7}

2) Now, we need to find triplets (x, y, z) that satisfy x > 2y > 3z

3) Let’s start with z:
– z must be small enough that 3z is less than some possible value of 2y
– The smallest possible value for y is 2 (since y > z)
– So, 3z < 2(2) = 4
– This means z can only be 1

4) Now for y:
– We know 2y > 3z = 3
– So y > 1.5
– Since y is a natural number, the smallest it can be is 2
– The largest it can be depends on x, but it can’t be larger than 3 (because 2(4) = 8, which is larger than any number in our set)

5) For x:
– x must be greater than 2y
– If y = 2, then x > 4
– If y = 3, then x > 6

6) Let’s list all possible triplets:
– When y = 2: (5, 2, 1), (6, 2, 1), (7, 2, 1)
– When y = 3: (7, 3, 1)

Therefore, there are 4 distinct triplets that satisfy the conditions.

The correct answer is d) Four triplets.