# How many five-digit numbers are there, if they contain three identical digits that are even number, and the remaining two digits the same odd number?

How many five-digit numbers are there, if they contain three identical digits that are even number, and the remaining two digits the same odd number?

If we don't count zero as a even number, I would calculate it this way: $$\binom{5}{1}\binom{4}{1}\cdot\frac{5!}{2!\cdot3!}$$. How could I calculate this if we take zero into account and if the number can't start with it?

• Welcome to MSE. A question should be written in such a way that it can be understood even by someone who did not read the title. Commented Nov 19, 2021 at 9:28
• Good start. Just carry on with the same technique with the remaining case where the even number is Zero. You have to start with Odd = X (how many ways?). Now, how many ways can you put "Zero Zero Zero X" into the last 4 digits? Commented Nov 19, 2021 at 9:35
• Formatting tip: Typing $\binom{n}{k}$ produces $\binom{n}{k}$. Commented Nov 19, 2021 at 10:21

So treat it as if you had only 4 digits available, three zeros and one odd digit to place ($$4*5$$). Once you calculate that, the sum of your calculations and this calculation is the answer.