How many four-digit odd numbers, all of digits different, can be formed from the digits 0 to 9, if there must be a 5 in the number? 
How many four-digit odd numbers, all of digits different, can be formed from the digits 0 to 9, if there must be a 5 in the number?

I know that there are 4 different cases where 5 is in the number:
5 _ _ _
_ 5 _ _
_ _ 5 _
_ _ _ 5
The first digit cannot be 0, since it has to be a 4-digit number.
 A: I would fill in the last number first and consider two cases: the last number is a $5$, and the last number is not a $5$.
Case 1: Last number is five.
This is the easy case, the remaining digits can be any distinct digits that are not $5$ (and do not begin with zero).  This gives $8$ choices for the first number (not 5 and not 0), $8$ choices for the second number (not the first number, and not $5$) and $7$ choices for the third number (not the first two, and not $5$--the last).  This gives $8\cdot 8\cdot 7=448 options.
Case 2: The last number is not five.
First, we will make sure the number is odd.  To do that, the last digit must be either $1, 3, 5, 7,$ or $9$.  Since it is not $5$, we have $4$ choices for the last digit.  Now, let us fill in the first digit (since it cannot be zero).  There are again $8$ choices for the first digit (not the last digit and not zero).  There are again, $8$ choices for the second number and $7$ for the third (for exactly the same reasons as mentioned before).  We now have $4\cdot 8\cdot 8\cdot 7=1792$ numbers.  However some of these numbers didn't use a $5$.  How many didn't use $5$?  Let's count that and subtract.  There are again $4$ choices for the last number (since $5$ wasn't an option there).  Now there are only $7$ choices for the first number: not $5$, not $0$, and not the last digit.  Similarly, there are $7$ choices for the second digit and $6$ choices for the third digit.  Thus, there are $7\cdot 7\cdot 6\cdot 4=1176$ numbers that don't use $5$ at all.  Thus we have $1792-1162=616$ good numbers.
Combining case 1 and case 2 gives us $448+616=1064$ four-digit, odd numbers with distinct digits and a $5$ occurring exactly once.
A: There are special conditions attached to the $\color{red}{\text{first}}$ and $\color{blue}{\text{last}}$ characters, so group into 3 cases:


*

*$5$ first: then for the remaining places there are (choosing the last digit first) $\color{blue}{4}\times 8\times 7 = 224$ options

*$5$ second/third: then for the remaining places there are $\color{blue}{4}\times \color{red}{7}\times 7 = 196$ options for each, so total of $2 \times 196 =392$ for this case.

*$5$ last: then for the remaining places there are $\color{red}{8}\times 8\times 7 = 448$ options


Total options then are: $224+392+448=1064$.
A: Another way to do this is to think of all the cases generally, and subtract all cases without any $5$s, because your question is essentially asking, "How many 4-digit odd numbers, all digits different, can we form where one of the digits is $5$?"
So, the number of 4-digit odd numbers in general is $8\cdot8\cdot7\cdot5$. Explanation: $5$ possible numbers in the last slot to make it odd, $8$ in the first slot because it can't be what was in the last slot and it can't be $0$, $8$ in the second slot (neither first nor last, but you can use $0$), and then $7$ in the third slot.
The number of 4-digit odd numbers without any $5$s is $7*7*5*4$. Explanation: $4$ possible numbers in the last slot to make it odd, sans $5$. Follow same arguments as above.
Final answer: $8\cdot8\cdot7\cdot5 - 7\cdot7\cdot5\cdot4=1064$.
