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