How many distinct numbers between $10$ and $1000$ can be formed from the digits $2$, $3$, $4$, $5$, $6$, and $0$ without repetition of any digit? Here is my attempt:

If $0$ is not to be any digit of the number, then there are $5 \times 4 \times 3 = 60 = ^5P_2$ possibilities for a 3-digit number, whereas there are $5 \times 4 = 20 = ^5P_3$ possibilities for a 2-digit number. So there are $60 + 20 = 80$ possible numbers between $10$ and $1000$ inclusive formed from the digits $2, 3, 4, 5, 6$.
If $0$ is to be in the unit place, then there are $5 \times 4 = 20$ possibilities for a 3-digit number, and there are $5$ possibilities for a $2$-digit number. So there are a total of $20 + 5 = 25$ numbers formed from these digits with $0$ in the unit place.
If $0$ is to be in the tenth place, then there are no 2-digit numbers, and $5 \times 4 = 20$ 3-digit numbers, totaling to $0 + 20 = 20$ numbers.
Thus there are a total of $80 + 25 + 20 = 125$ numbers.

Is this calculation correct?
If not, then where lies the problem?

First we deal with all two digit numbers ($11 - 99$). There are two numbers to pick, and tens place cannot be $0$. Hence number of ways are $5\cdot 5$.
Now three digit numbers ($100-999$). Again we cannot have $0$ at hundreds place, so we have $5\cdot 5 \cdot 4$ ways.
In all we have $5^3$ numbers.
