**Concept: The question tests on the area of Counting**

**Solution:**

If the first digit is 1, then only zero is less than 1. We can have 0 in the units and 10's place

Therefore we can have 1 * 1 = 1 number (100)

If 2 is in the first place, then 2 numbers 0 and 1 are possible.

We can have 2 * 2 = 4 numbers (200, 211, 201, 210)

If 3 is in the hundreds place, then 0, 1 and 2 are lesser and we can have 3 * 3 = 9 numbers

If 9 is in the units place, we will have 9 numbers 0 to 8 which are less than 9, and we can have 9 * 9 = 81 numbers.

So the sequence is the sum of squares from 1 till 9 = n(n+1)(2n+1)/6

=9∗10∗19/6

=285

(**option a)**