A number will be divisible by 6 iff the digit at the unit place of the number is divisible by 2 and sum of all digits of the number is divisible by 3 .

Units, place must be occupied by 4 and hence, at least one 4 must be there.

Possible combination of 4, 5, 9 are as follows :

4	5	9	Total number of Number
1	1	4	$${{5!} \over {4!}} = 5$$
1	4	1	$${{5!} \over {4!}} = 5$$
2	2	2	$${{5!} \over {2!2!}} = 30$$
3	0	3	$${{5!} \over {2!3!}} = 10$$
3	3	0	$${{5!} \over {2!3!}} = 10$$
4	1	1	$${{5!} \over {3!}} = 20$$
6	0	0	$${{5!} \over {5!}} = 1$$

Total = 5 + 5 + 30 + 10 + 10 + 20 + 1 = 81.