How many four digit positive integers, which are divisible by 6, can b

The sets are {2,3,4,6}, {0,2,4,6}, {0,2,3,4}

If you add up the numbers in individual sets, you get a number that is divisible by 3. No other set of 4 digits from the 5-digit set provided fulfils that criterion.

I just realized this post is from 2020. Hope it still helps future readers.

* To arrive at 4 set of possibilities.*

First, find the sum of the given 5 digits to get number 15. This 15 is divisible by 3. So, number with 4 digits which are divible by 3 must also exclude a digit which is divisible by 3. So, three possiblities we get by excluding 0,3 and 6.

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­Number will be div by 6 i.e. the number will be div by both 2 and 3.
Conditions:
1. For div by 2, the resultant will be an even number.
2. For div by 3, the sum of digits is div by 3.

With given digits, we can get the sum 9, 12 and 15 using these combinations: (0,2,3,4), (0,2,4.6) and (2,3,4,6).

(0,2,3,4): There will be 2 ways to arrange the digits.
1. when the unit is 0 then, 
unit is filled in 1 way (only 0), thousands in 3 ways (2,3 or 4), hundreds in 2 ways (after selecting any one from 2,3,4 in the thousands place, select one of the remaining two) and tens in the remaining 1 way. So 3*2*1*1=6 ways.
2. when the unit is not 0 then,
unit is filled in 3 ways (since we have to create an even number so select one from 0, 2 or 4), thousands in 2 ways (the remaining one from the unit which is not 0 and 3), hundreds in 2 ways (remaining two from thousands) and tens in 1 way. So 2*2*1*2=8 ways.
Total 6+8=14 ways.

(0,2,4,6): Follow the above cases and we will get 6 ways (when the unit is 0) and 12 ways (when the unit is not 0). So total 6+12=18 ways.

(2,3,4,6): unit in 3 ways (either 2,4 or 6), thousands in the remaining 3 ways, hundreds in 2 ways and tens in 1 way. So 3*2*1*3=18 ways.

Total: 14+18+18=50. Option (D) is correct.