How many four digit numbers, which are divisible by 6, can be formed using the digits 0, 2, 3, 4, 6, such that no digit is used more than once and 0 does not occur in the left-most position?

50

From CAT 2017 – Quantitative Aptitude – Modern Maths – Permutation and Combination, we can see that,

For a number to be divisible by 6, it should have both 2 and 3 as factors.

For 3: The sum of digits should be divisible by 3

For 2: The last digit of the number should be even

There can be three such cases, (0,2,3,4), (0,2,4,6) and (2,3,4,6)

1st case: (0,2,3,4)

When the last digit is â€˜0â€™

The number of combinations can be 3! = 6

When the last digit is 2/4

The number of combinations will be 2*2*1*2 = 8

Total = 14

2nd case: (0,2,4,6)

When the last digit is â€˜0â€™

The number of combinations = 3! = 6

When the last digit is 2/4/6

The number of combinations = 2*2*1*3 = 12

Total = 18

3rd case: (2,3,4,6)

Last digit has to be 2/4/6

So, the number of combinations = 3*2*1*3 = 18

Total number of combinations = 14+18+18 = 50

Answer: 50

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