How many distinct odd-digit numbers can be written with the digits 1,2,3, 4 and 5 if no digit may be used more than once?
+182 There are some great explanations out there. Start checking them out before you post an question
https://web2.0calc.com/questions/help_16304 (No credit to me from the link)
1, 2, 3, 4, 5 ==5 digits
5! ==120 permutations. Each permutation begins with one of the 5 digits this many times: 120 / 5 ==24 permutations. How many odd numers are there? There are 3 such numbers: 1, 3 and 5.
Therefore, the total number of ODD permutations ==3 x 24 ==72 such permutatins.
