Question

What is the probability that a permutation of 2n has each of 1, . . . , n appearing before n +1,...,2n?

Answer 1

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Since there are total of 2n numbers

So, to permute total 2n numbers it is (2n)!

To permute first n numbers in any order, it is n!

To permute last n numbers in any order, it is n!

So, Let us say there are 2 bunch with first n and last n numbers

So, first bunch will always be before second

So, total cases will be n!*n! to arrange first n before last n

So, the probability is

P=n!*n!/(2n)!

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