Question

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

Answer #1

Hey,

**Note: Brother in case of any queries, just comment in
box I would be very happy to assist all your queries**

**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)!**

**Kindly revert for any queries**

**Thanks.**

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