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Question

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answer is 8

what approach did u use ??

i was doing like

numbers div by 2 = 4*3*2=24

numbers div by 3= 4* 3 *1=12

but i was not able to find the numbers which will be divisible by both 2 and 3 and hence divisible by 6

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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

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thanks!