number divisibility count

For a number that is divisible by 6, it must be an even number and the sum of its digits must be divisible by 3, this is because 2 and 3 are prime factors of 6. To be an even number the last digit must be either 2, 4 or 6. The sum of digits 1+2+3+4+5+6=21, no matter how you arrange it is 21 and divisible by 3. The possibility of a number with last digit 2, 4 or 6 is 3×5! ÷ 6! = 360/720 =0.5 Voila!

All six digits number with 1,2,3,4,5,6 is divisible by 3.

It's divisible by 6 if it's an even number.

So the ones digit can only be 2,4,6. It can't be 1,3,5.

Probability=3/6= 0.5

In order for a number to be divisible by 6, it must have 6, 4 and 2 in the unit place. The total number of arrangements having these numbers in the unit place is 5!+5!+5!= 360. The total number of numbers having 6 digits in their random order is 6!= 720. The probability is 360/720= 1/2 = 0.5.