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?


That is absolutely correct. Here’s another way to think about the digits:

We need to start our number with something that isn’t zero. There are 5 possibilities.

We need a next digit and it might be zero but it can’t be the same as the previous so there are 5 possibilities

The third digit can be any digit we’ve not already used, including zero, (4 possibilities) or it can be “no digit at all” (ie end the number at two digits). Thus there are 5 possibilities

So the answer is $5\times5\times5=125.$

We note that we don’t care about 4 digit numbers because the only 4 digit number to consider is 1000 and that has a 1 in it which we aren’t allowed.


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.

  • $\begingroup$ @B.Goddard Thats what we get without repetition only. tens place has 5 possibilities, and ones place too has 5 (4 unused + "0") $\endgroup$ – samjoe Dec 13 '17 at 12:17

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