A strategy for a basic combinatorics problem Given the set $A=\left\{2;\:3;\:4;\:5;\:6;\:7;\:8;\:9\right\}$, how many five digit, odd, distinct, numbers can we construct with the elements of $A$, that are also less than $49999$.
Here's how I tried to solved it:

If $9$ is in the units, 2 or 3 must be in the tenths of thousands. Two numbers selected out of eight, there remains six numbers for three places (between the units and the tens of thousands). Therefore: $2\cdot_6P_3$
If $3$ is in the units place, two choices remain for the tens of thousands. Again: $2\cdot_6P_3$
If $5$ or $7$ are in the units, there can be three numbers in the tens of thousands ($2$, $3$, or $4$). Then, there are still six numbers remaining to chose. Therefore: $2\cdot3\cdot_6P_3$
The number of combinations shoulb be given by $2\cdot_6P_3+2\cdot_6P_3+2\cdot3\cdot_6P_3=1200$

This solution is wrong, and I can figure out why.
 A: I assume that you are not allowed to repeat digits in the number: for example $33333$ doesn't count as a valid odd five-digit number less than $49999$ made from the available digits.

We know that the first digit (in the ten-thousands place) must be a $2,3,$ or $4$ in order to satisfy the condition that the result is less than $49999$.  Break into cases.
Once having done so, pick the final digit next (in the units place).  It must be an odd digit from those remaining (note that if $3$ was selected for the first digit then this affects the count)
Once having done so, finally fill in the remaining digits from left to right with any still available to choose from.


*

*If the first digit is even, i.e. it is a $2$ or a $4$, pick which one it is  (2 options)


*

*The final digit is odd, pick which it is (4 options)

*Pick the second, third, and fourth digits (6, 5, and 4 options each respectively)


*If the first digit was odd, then it must have been a $3$ (1 option)


*

*The final digit is odd, pick which it is (3 options)

*Pick the second, third, and fourth digits (6, 5, and 4 options each respectively)



This gives a total of $2\times 4\times 6\times 5\times 4 + 1\times 3\times 6\times 5\times 4 = 1320$

Your error appears to be in the line "If $9$ is in the units, 2 or 3 must be in the tenths of thousands"
Note that the number $45679$ is a valid number which ends in a nine and yet begins with a $4$.  You neglected to take $4$ into consideration as a possible leading digit when $9$ is the final digit.
A: You considered all $4$ cases on the units (last) digit being odd.
Alternatively, you can consider $2$ cases ($[\ ]$ is the specific digit of the $5$-digit number):
Case 1: $[2 \ or \ 4][\  ][\ ][\ ][3] \Rightarrow 2\cdot P(6,3)=2\cdot 120=240;$
Case 2: $[2 \ or \ 3 \ or \ 4][\  ][\ ][\ ][5 \ or \ 7 \ or \ 9] \Rightarrow 3\cdot 3\cdot P(6,3)=3\cdot 3\cdot 120=1080.$
Hence, the number of favorable numbers is: $240+1080=1320$.
Yes, as noted by JMoravitz, you missed $\color{red}4[\ ][\ ][\ ][9]$ in the first case: hence, you must have $\color{red}3\cdot_6P_3$.
