+0

# Counting

0
125
5

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?

Jul 8, 2022

#1
+2541
0

Note that there are 5 choices for the final digit.

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

Jul 8, 2022
#4
+2541
0

BuilderBoi  Jul 9, 2022
#2
+285
-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.

Jul 8, 2022
edited by Voldemort  Jul 8, 2022
#5
+285
-1

Disregard my message as well.

Voldemort  Jul 9, 2022
#3
+1

Distinct odd-digit numbers. It means only numbers that comprise of odd digits only.

[(1,), (3,), (5,)] [(1, 3), (1, 5), (3, 1), (3, 5), (5, 1), (5, 3)] [(1, 3, 5), (1, 5, 3), (3, 1, 5), (3, 5, 1), (5, 1, 3), (5, 3, 1)] >>>Total distinct permutations with odd digits only == 15

Jul 8, 2022