The correct option is
B 154Here, the available digits are
0,1,2,3,4,5,6,7,8,9.The numbers can be of one, two or three digits and in each of them unit's place must have 0 or 5 as they must be divisible by 5.
The number of numbers of one digit =1
(∵ is the only number).
The number of numbers of two digits divisible by 5= number of all the numbers of two digits divisible by 5− number of numbers of two digits divisible by 5 and having 0 in ten's place =2P1×9P1−1,
(∵unit's place can be filled by either 0 or 5 in first category and only by 5 in the second category)
=2×9−1=17.
The number of numbers of numbers of three digits divisible by 5= number of all the numbers of three digits divisible by 5= number of numbers of three digits divisible by 5= number of numbers of three digits divisible by 5 and having 0 in hundred's place
=2P1×9P2×8P1×1=2×9×8−8=136.
∴ required number of numbers
=1+17+136=154