Permutations and Combinations Part 2

A number is multiple of 15 when the number is divisible by 5 and sum of digits of the number is divisible by 3.
Among 1, 2, 3, 4, 5, 6, 7 unit place is filled with 5 so it is multiple of 5.
Now to make it divisible by 3, take remaining 5 digits such a way that sum becomes divisible by 3.
Remaining 5 digits can be
(1) 1, 2, 3, 4, 6
Here sum = 1 + 2 + 3 + 4 + 6 + 5 = 21 (divisible by 3)
This 5 digits can be filled in those 5 placed without repetition in 5 4 3 2 1 = 51 = 120 ways
(2) 2, 3, 4, 6, 7
Here sum = 2 + 3 + 4 + 6 + 7 + 5 = 27 (divisible by 3)
∴ Number of ways =51=120
(3) 1,2,3,6,7
Here sum =1+2+3+6+7+5=24 (divisible by 3 )
∴ Number of ways =51=120
∴ Total possible 6 digit numbers divisible by 15
=120+120+120=360