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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
avatar+31 
+1

The answer would be: 

m x n x p = 5 x 4 x 3  = 60

So the answer would be 60 

 Jun 10, 2022
 #2
avatar+2448 
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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
avatar+124676 
+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

 

 

cool cool cool

 Jun 11, 2022

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