How many 3-digit positive integers are odd and do not contain digit 5?

There are four digits that the number can end with and be odd, not $\frac{9}{2}$, which is what your calculation uses -- that is, there are more even numbers without a five than odd numbers without a five.

More correctly:

$8 * 9 * 4 = 72 * 4 = 288$, as the first digit can be any of $1,2,3,4,6,7,8,9$, the second any but $5$, and the third must be $1,3,7,$ or $9$.


There is no reason that there are just as many odd integers that do not contain $5$ as there are even integers that do contain 5. The proper fraction is $\dfrac{4}{9}$.

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Permutations

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