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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?

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    $\begingroup$ 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. $\endgroup$ Commented Nov 19, 2021 at 9:28
  • $\begingroup$ 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? $\endgroup$ Commented Nov 19, 2021 at 9:35
  • $\begingroup$ Formatting tip: Typing $\binom{n}{k}$ produces $\binom{n}{k}$. $\endgroup$ Commented Nov 19, 2021 at 10:21

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That calculation is correct. Now if you take zero into consideration, all you have to do is notice, that zero cannot be the most significant digit. So the zero can only be placed on the 4 other digits -> the most significant digit has to be odd.

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.

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