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:

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> 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: $2\cdot\_6P\_3$
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> If $3$ is in the units place, two choices remain for the tens of thousands. Again: $2\cdot\_6P\_3$
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> 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: $2\cdot3\cdot\_6P\_3$
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> The number of combinations shoulb be given by $2\cdot\_6P\_3+2\cdot\_6P\_3+2\cdot3\cdot\_6P\_3=1200$
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> 

This solution is wrong, and I can figure out why.