Let's think about how you'd do it in base ten, and you can apply it to base 6.
One way is to think digit by digit, as you have apparently started to do. If the units digit is zero, the first two digits can be anything from 1 to 15, so there are 15. If the tens digit is zero, the units digit can be anything from 1 to 9 (rejecting 0 because that would be already counted), and the hundreds digit has to be 1 (because if it were zero, we wouldn't write the zero in the tens place). This makes 9 more. Finally, if the hundreds digit is 0 ... we wouldn't write the zero, so this adds none. Total: 15+9=24.
To check, here are the numbers:
10, 20, 30, 40, 50, 60, 70, 80, 90, 100, 110, 120, 130, 140, 150,
101, 102, 103, 104, 105, 106, 107, 108, 109
I don't know whether that counts in your mind as "using p&c". This problem is not one of permutations or combinations in the technical sense, since digits can be repeated. But in terms of combinatorics more generally, my first thought would be to subtract the number of numbers that can be made
without zero, from the total of 150 numbers. I'll leave you to think about how to do that; it won't necessarily be any easier than the first way, because the number of possibilities in one place depends on the others. I'd break the interval from 1 to 150 into segments and count each separately. (I do get the same answer both ways.)
