Correct Option
The numbers x, y, and z are selected from the set of the first seven natural numbers, i.e., {1, 2, 3, 4, 5, 6, 7}. The given condition is x > 2y > 3z.
To determine the possible values for z:
- Since z is a natural number, z ≥ 1.
- If z = 2, then 3z = 6. The condition becomes x > 2y > 6.
- From 2y > 6, it implies y > 3. The smallest possible integer value for y from the given set would be 4.
- If y = 4, then 2y = 8. The condition would then require x > 8.
- However, x must be selected from the set {1, 2, 3, 4, 5, 6, 7}. No value in this set satisfies x > 8.
- Therefore, z cannot be 2 or any value greater than 2.
- The only possible value for z is 1.
Now, with z = 1, the condition simplifies to x > 2y > 3.
To determine the possible values for y:
- From 2y > 3, it implies y > 1.5.
- Since y is a natural number from the set {1, 2, 3, 4, 5, 6, 7}, possible values for y are {2, 3, 4, 5, 6, 7}.
Let's evaluate the possible triplets (x, y, z) based on these constraints:
- Case 1: y = 2
- 2y = 4. The condition becomes x > 4.
- Possible values for x from {1..7} that satisfy x > 4 are {5, 6, 7}.
- This yields the triplets: (5, 2, 1), (6, 2, 1), (7, 2, 1). (3 triplets)
- Case 2: y = 3
- 2y = 6. The condition becomes x > 6.
- Possible values for x from {1..7} that satisfy x > 6 are {7}.
- This yields the triplet: (7, 3, 1). (1 triplet)
- Case 3: y = 4
- 2y = 8. The condition becomes x > 8.
- No value of x from {1..7} satisfies x > 8.
- Case 4: y ≥ 4
- For any y ≥ 4, 2y will be 8 or greater. Consequently, x > 2y would imply x > 8, which is not possible within the given set of natural numbers {1..7}.
The total number of distinct triplets (x, y, z) possible is the sum of triplets from Case 1 and Case 2: 3 + 1 = 4.
Incorrect Options:
Options 1, 2, and 3 are incorrect because a systematic evaluation of the given condition x > 2y > 3z, with x, y, z selected from the first seven natural numbers, yields exactly four distinct triplets. The derivation demonstrates that only z=1 is possible, and subsequently, only specific values of y (2 and 3) allow for valid x values, resulting in a total of four such triplets.