How many different permutations are there of the numbers 1, 2, 3, 4, 5 so that...

How many different permutations are there of the numbers 1, 2, 3, 4, 5 so that the even numbers and the odd numbers alternate in the list?

Expert Solution

There are 5 numbers.So according to the given condition for 1st place of the 5 digit no we have 3 option of odd numbers 1,3,5.Now for the 2nd place 2 options as there are 2 even numbers 2 and 4.Now for the 3rd place we have 2 option numbers except one which came in 1st place.Then for 4th place only have 1 option.And for 5th place we have 1 option.

So total no of ways= 3*2*2*1*1=12

Now in the previous case we asumed that the number started with 1st digit odd

There will be another case where 1st digit will be even.And then we will find another 12 cases.

So total no of all possible permutations is= 2*12=24

orchestra answered 2 years ago

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