Explanation:

Let’s break down the problem:

- We need to form 6-digit numbers using 0, 1, 2, 3, 4, 5 exactly once each.

- The number must be divisible by 6, so:

- Must be divisible by 2 ? last digit even (0, 2, 4 are options).

- Must be divisible by 3 ? sum of digits is divisible by 3.

- The total sum of digits is 0+1+2+3+4+5 = 15, divisible by 3 . So, all permutations will be divisible by 3.

- For divisibility by 2, last digit must be 0, 2, or 4.

Let’s count:

- Case 1: Last digit = 0

- First digit cannot be 0.

- Remaining digits: 1,2,3,4,5 ? Any can be first, permutations = 5! = 120

- Case 2: Last digit = 2

- Remaining digits: 0,1,3,4,5

- First digit can’t be 0 or 2 (so can be 1,3,4,5) ? 4 options.

- Arrange the rest: 4! = 24

- So, 4 × 24 = 96

- Case 3: Last digit = 4

- Remaining digits: 0,1,2,3,5

- First digit can’t be 0 or 4 (so can be 1,2,3,5) ? 4 options.

- Arrange the rest: 4! = 24

- So, 4 × 24 = 96

- Total: 120 + 96 + 96 = 312

Review of the options:

- Option 1: 96 ? (under-counts)

- Option 2: 120 ? (considers only one case)

- Option 3: 192 ? (too low)

- Option 4: 312 (This is correct)

Option 4: 312 is correct.