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?
+2666 Note that there are 5 choices for the final digit.
Of these, 2 are odd, so the probability is \(\color{brown}\boxed{2 \over 5}\)
+288 I believe the problem is asking for odd-digit numbers and not odd numbers. So we have to consider three cases:
1. 1 digit numbers
There are 5 1 digit numbers that may be formed.
2. 3 digit numbers
There are 5 x 4 x 3 = 60 3 digit numbers.
3. 5 digit numbers.
There are 5 x 4 x 3 x 2 x 1 = 120 5 digit numbers.
Adding all of these up gives us 185.
