I need to find the number of $10$ digit numbers containing all odd digits (other digits are allowed, but the number must contain $1$, $3$, $5$, $7$ and $9$). Also, this number cannot contain a $0$.

I have come up with this solution but I am not sure if it's correct:
First, we choose the place-holders for those odd digits - we can do this in ${10 \choose 5} $ ways. Now, arrange our odd digits in these placeholders - $5!$ ways. Next, we fill in the gaps - every digit can go there, and so we have $9^5$ ways to do this. Therefore, the answer is
$${10 \choose 5 } 5! \cdot9^5$$ Is this correct?
Now, how could I solve the exact same problem if we allowed $0$ to appear in this number? Namely, I am addressing the problem of $0$ as the first digit.

  • 2
    $\begingroup$ Your method multiply counts numbers with more than one $1$, for example. $\endgroup$ – lulu Mar 12 '18 at 0:48
  • $\begingroup$ @lulu but it is allowed right? 'Next, we fill in the gaps - every digit can go there'. Sorry for bad english $\endgroup$ – Gareth Ma Mar 12 '18 at 0:49
  • 1
    $\begingroup$ @KGSHbteamMineTeamBeastO_ Of course you are allowed to have more than one $1$ but you can't count the same number twice, that's the problem. $\endgroup$ – lulu Mar 12 '18 at 0:50

It is not correct because you counted $1135792468$ twice, once when the first $1$ was among your first five numbers and once when the second $1$ was among your first five numbers.

You want to use the inclusion-exclusion principle. There are $9^{10}$ ten digit numbers with no zeros. Subtract the ones that are missing $1$, those that are missing $3$, etc. You have subtracted the ones missing both $1$ and $3$ twice, so add them back in and so on. If you search the site there are many examples.


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