The no should be divisible by both 2 and 3 to be divisible by 6 so for this
Last digit of the no should be either 2 or 4
Lets first take last digit be 2 then the other 2 digit should sum to either 4 or 7 Why? Because sum of all 3 should be divisible by 3 so for this possible value are for 4-(1,3) can't take (2,2) as repetition not allowed so possible 3 digit no are-132 or 312 similarly for 7 (4,3) and 2 permutation 432 and 342.
Similarly take 4 in unit place and the other two digit sum can be 5(2,3) or 8(5,3)
So total 8