Question

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

Answer 1

We have 0 + 2 + 3 + 4 + 6 = 15.
Since number is to be divisible by 6 meaning it need to be divisible by 2 and 3.
So, sum of digits need to be multiple of 3 and unit digit should be an even number.
Possible combination:
1. 2, 3, 4, 6
2. 0, 2, 3, 4
3. 0, 2, 4, 6
For 1st combination: Units place can be taken by 3 numbers, tens can be taken by 3, hundreds by 2 and thousand s place by 1 way.
Numbers possible in this case = 1 x 2 x 3 x 3 = 18
For 2nd combination: Total number of numbers possible without any condition = 4! = 24.
Now out of 24, 6 will have 3 at units place and hence to be eliminated.
6 numbers will have 0 at thousands place, hence need to be eliminated.
Out of 12 eliminated numbers, 2 numbers which are 0243 and 0423 are deleted twice and hence need to be added.
Therefore, numbers possible in this cases = 24 – 6 – 6 + 2 = 14.
For 3rd combination: Total number of numbers possible without any condition = 4! = 24.
Out of 24, 6 will have 0 at thousands place and hence need to be eliminated.
Therefore, numbers possible in this cases = 24 – 6 = 18
Total numbers possible = 18+ 14 + 18 = 50.