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?

Guest Jul 8, 2022

#1**0 **

Note that there are 5 choices for the final digit.

Of these, 2 are odd, so the probability is \(\color{brown}\boxed{2 \over 5}\)

BuilderBoi Jul 8, 2022

#2**-1 **

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.

Voldemort Jul 8, 2022