Always choose the most restrictive digit first.  Then choose the next most
restrictive, etc.

The most restrictive digit is the 5th because that's the one that 
determines that it is an odd number.  So we will choose the fifth 
one first.

The next most restrictive digit is the first one, since it cannot
be 0.  So we will chose it second.

We can choose the 5th digit any of 5 ways, {1,3,5,7,9}

That's 5 ways to choose the 5th digit. Next we will choose the 1st
digit.

Choosing the 5th digit leaves 9 digits not chosen.  However one of 
those is 0, since we did not choose 0 for the 5th digit.  And we 
cannot choose 0 for the first digit, so there are only 8 choices
for the 1st digit. 

So there are 5×8 or 40 ways to choose the 5th and 1st digits.

For each of those 5×8 or 40 ways to choose the 5th and 1st digits,
there are 8 digits left to choose for the 2nd digit. (it CAN be 0).

So there are 5×8×8 or 320 ways to choose the 5th, 1st and 2nd digits.

For each of those 5×8×8 or 320 ways to choose the 5th, 1st, and 2nd
digits, there are 7 digits left to choose for the 3rd digit.

So there are 5×8×8×7 or 2240 ways to choose the 5th, 1st, 2nd, and 3rd
digits.      

For each of those 5×8×8×7 or 2240 ways to choose the 5th, 1st, 2nd, and
3rd digits, there are 6 digits left to choose for the 4th and final
digit to choose.

So there are 5×8×8×7×6 or 13440 ways to choose the 5th, 1st, 2nd, 3rd,
and 4th digits, with no digits repeated, which is the answer to the
problem.

Edwin