Permutations and Combinations Part 1

Given, digits ={1,2,3,4,5}
Numbers divisible by 3 (sum of digits divisible by 3 ).Case I When sum is 12→3,4,5→3!=6
Case II When sum is 9→2,3,4→3!=6
Case III When sum is 9→1,3,5→3!=6
Case IV When sum is 6→1,2,3→3!=6
So, total numbers divisible by 3=6×4=24
Numbers divisible by 5 (ending with 5)
So, total numbers divisible by 5=12
Numbers divisible by 15 , are 145,415,345,435
i.e. total 4 numbers are divisible by both 3 and 5. i.e. divisible by 15 .
Hence, the required numbers which are divisible by 3 or 5 =24+12−4=32