How many odd integers are there between $0$ and $100$ that have distinct digits? I'm interested in counting the odd integers between $0$ and $100$ with distinct digits using a combinatorial approach.
My attempt is as follows. There are $10$ possible choices for the first digit, that is the tens digit. Then, the second digit can be chosen in $5$ ways if the tens digit is even, and in $4$ ways if the tens digit is odd. But then I'm stuck since the two events seem to be dependent and I don't see a straightforward way to apply the multiplication rule.
 A: You have the think the other way around.
First, let's compute the number of odd integers between 0 and 99. We can leave out 100 since it is even. Any integer between 0 and 99 has a unit digit and a tens digit, i.e. write
$1,2,\ldots 9$ as $00, 01, 02,\ldots 09$. Let $U$ be the event of choosing a digit for the unit. This can be done in 5
ways (e.g. 1,3,5,7 or 9). Then let $D$ be the event of choosing a digit for the tens; this can be done in 10 ways. Since the number of $U$ that can occur doesn't depend on the number of $D$ that can occur, and vice versa, the required number is $10\cdot 5=50$.
Now back to your problem. Let $D$ and $U$ be as above. Then $U$ can occur again in 5 ways and after this,
$D$ can occur in $9$ ways. This is because if $U$ happens to be 1, then $D$ should be anything than 1. Since the number of $U$ that can occur doesn't depend on the number of $D$, the multiplication rule applies and the result is $9\cdot 5=45$.
A: I would think in the "opposite" direction. You have 5 possibilities for the second digit, and then 9 possibilities for the first one: 45 numbers.
Using your approach, you have $5\cdot 5+ 5\cdot 4 =45 $ possibilities.
Let $S$ be the set of numbers you are interested in.
$5\cdot 5$ is the number of elements of $S$, whose first digit is even; while $5\cdot 4$ is the number of elements of $S$, whose first digit is odd.
The union of these 2 subsets is the whole $S$, and since they are disjoint, it holds $|S|= 25+20$.
