# How many odd numbers of $5$ digits can be formed with the digits $0,2,3,4,5$ without repetition of any digit?

How many odd numbers of $5$ digits can be formed with the digits $0,2,3,4,5$ without repetition of any digit?

I noticed that the last number can be filled in $2$ ways and first place can be filled in $3$ ways. Consequently, 2nd 3rd and 4th places can be filled in $3$, $2$ and $1$ ways respectively. Is there any flaw in this reasoning? Please guide further.

• $\color{green}\checkmark\quad$ You are good to go. – Graham Kemp Jul 13 '17 at 5:21
• I will warn you about naming things "permutations" but then treating them as different objects. The phrasing of the question in the body is correct and unambiguous ("how many odd numbers of 5 digits..."), but the concept of an "odd permutation" is distinctly different than that of an odd number. For example, 52340 is an odd permutation of the digits when treated as a permutation of $\{0,2,3,4,5\}$ despite being an even number. If in doubt of whether something should be referred to as a permutation, "arrangement" is a more generic term which is usually valid and often more appropriate. – JMoravitz Jul 13 '17 at 5:44
• ...(odd permutation here in the algebraic sense as in the permutation can be expressed as the product of an odd number of transpositions) – JMoravitz Jul 13 '17 at 5:46

$\color{green}\checkmark\quad$ You are good to go.
We count ways to select from the two odd digits for the units, from the three other non-zero digits for ten-thousandth place, and how to arrange the three remaining digits in the remaining places.   Then we go forth and mulltiply. $$2\cdot 3\cdot 3! ~=~ 36$$