+0

# Counting

+1
34
3

How many three digit odd numbers can be formed from the digits 1, 2, 3, 4 and 5, if no digits are to be repeated?

Jun 10, 2022

#1
+27
+1

m x n x p = 5 x 4 x 3  = 60

So the answer would be 60

Jun 10, 2022
#2
+1746
+1

There are $$5 \times 4 \times 3 = 60$$ ways to choose without any restrictions.

Note that of these integers, exactly $$3 \over 5$$th of them are odd because there are 5 choices for the final digit, 3 of which, are odd.

This means that there is $${3 \over 5} \times 60 = \color{brown}\boxed{36}$$ numbers.

Jun 11, 2022
#3
+123309
+1

We have three choices for a final digit

And  for each of these,  we can  permute  any 2 of the 4  remaining digits

So

3 * P( 4,2) =  3 * 12   =  36

Jun 11, 2022