permutation sum

[winnerhere]

The first n natural numbers , 1 to n, have to be arranged in a row from left to right. The n numbers are arranged such that thee are an odd number of numbers between any two even numbers as well as between any two odd numbers. If the number of ways in which this can be done is 72, then find the value of n

1) 6

2) 7

3) 8

4) 9

5) 10

Pleas ehlp to udnerstand the logic behing solving this sum

Thanks,
Sai

[rkanthilal]

There are n numbers from 1 to n. There is an odd number of numbers between any two even numbers as well as between any two odd numbers. This basically means that the numbers are ordered by alternating between odds and evens and that the number of odds and evens are equal or in the case of an odd n value there will be one more odd or even number. Any other way would violate this condition.

Working through the answers starting with 6. The question becomes if you have three odd numbers and three even numbers, can they be arranged in 72 different ways while meeting the condition of alternating between odds and evens. There are two ways to start the arrangements. You can start with an odd number or an even number. So you would have E-O-E-O-E-O or O-E-O-E-O-E. The three even numbers can be arranged in 6 ways (3!) and the three odd numbers can also be arranged in 6 ways (3!). By multiplying these two together you get the total number of ways you can arrange 3 odd number and 3 even numbers. That means there are 36 different ways to arrange the numbers if you start with an even number and there are another 36 different arrangements if you start with an odd number. This gives you a total of 72 arrangements.

I'm not sure if this is correct but I would choose n=6. Do you know what the answer is?

[winnerhere]

rkanthilal,

6 is a right answer

Thanks for the explanation!

Regards,
Sai