Solution:

To form a number divisible by 6, it must be divisible by both 2 and 3.

Divisibility by 2:

For the number to be divisible by 2, the last digit must be even i.e. 0, 2, 4 or 6.

Divisibility by 3:

For the number to be divisible by 3, the sum of all digits must be divisible by 3.

Let's consider the digits {0, 2, 3, 4, 6}.

Sum of all digits:

The sum of all digits is 0 + 2 + 3 + 4 + 6 = 15, which is divisible by 3.

Last digit:

The last digit can be either 0, 2, 4 or 6.

Thousands digit:

Since 0 cannot occur in the left-most position, the thousands digit can be any of the remaining 4 digits {2, 3, 4, 6}.

Hundreds digit:

The hundreds digit can be any of the remaining 3 digits {2, 3, 4, 6}.

Tens digit:

The tens digit can be any of the remaining 2 digits {2, 3, 4, 6}.

Total number of ways:

Using the multiplication principle, the total number of ways to form a four-digit number divisible by 6 is:

4 × 4 × 3 × 2 = 96

However, we need to eliminate the numbers that have repeated digits. There are 4 ways to choose the repeated digit and 3 ways to choose its position. Therefore, the total number of numbers with repeated digits is:

4 × 3 = 12

Hence, the total number of four-digit numbers divisible by 6 with no repeated digits is:

96 - 12 = 84

However, we need to eliminate the numbers that have 0 in the left-most position. There are 3 ways to choose the thousands digit (2, 3, 4 or 6) and 3 ways to choose the hundreds digit (excluding the digit already chosen for the thousands digit). For each choice of thousands and hundreds digit, there are 2 ways to choose the tens digit and 1 way to choose the units digit (since 0 is not allowed). Therefore, the total number of four-digit numbers divisible by 6 with no repeated digits and no 0 in the left-most position is:

3 × 3 × 2 × 1 = 18

Hence, the required number of four-digit numbers is:

84 - 18 = 66

But, we need to consider only those numbers which have 0 as the last digit. There are 3 ways to choose the thousands digit (2, 3, 4 or 6) and 3 ways to choose the hundreds digit (excluding the digit already chosen for the thousands digit). For each choice of thousands and hundreds digit, there is 1 way to choose the tens digit and 1 way to choose the units digit (since 0 is the last digit). Therefore, the total number of four-digit numbers divisible by 6 with no repeated digits, no 0 in the left-most position