Solveeit Logo
Login

Question

Question: In how many ways can \(2n\) people be seated, \(n\) at a round table and \(n\) in a row?...

In how many ways can 2n2n people be seated, nn at a round table and nn in a row?

Explanation

Solution

We will first calculate the permutations of nn people sitting around a circle and then calculate the permutations of nn people sitting in a row and then combine both of them to get the final answer.

Formula used: Combinations = nCr = n!(n - r)!{\text{Combinations = }}{}^{\text{n}}{{\text{C}}_{\text{r}}}{\text{ = n!(n - r)!}}

Complete step-by-step solution:
We know that there are a total 2n2n numbers of people which have to be seated.
There have to be nn number of people selected out of the total 2n2n number of people, this can be found using: 2nCn{}^{2n}{C_n}
This can be simplified by using the formula as: (2n!)(2n−n)!(2n!)(2n - n)!
On subtracting the bracket terms we get, (2n!)(n)!(2n!)(n)!
Now for the first nn which have to be seated at a round table:
The first person out of the nn people will have total nn places to be seated at the table,
Now since it is a circle, the other choices collapse because in the last permutation, it will become equal to another combination of people.
Therefore, the number of ways in which nn people can be seated round a table is:
(n−1)×(n−2)....×1(n - 1) \times (n - 2).... \times 1
This could be also represented as (n−1)!(n - 1)!.
Now the total numbers of ways nn items can be arranged are in n!n! ways because it is a row and the choice won't collapse after the first person has been seated.
Now the total number of ways 2n2n number of people can be seated will be:
Selection of nn people out of 2n×n{{2n}} \times {\text{n}} people sitting in a circle ×n{{ \times n}} people sitting in a row
This could be written as:
=(2n!)×n!×(n−1)!×n!(2n!) \times n! \times (n - 1)! \times n!
On multiplying both we get:
=2n!(n!)2(n−1)!2n!{(n!)^2}(n - 1)!

The required answer is 2n!(n!)2(n−1)!2n!{(n!)^2}(n - 1)!

Note: In these types of questions the formula for permutations and combinations should be memorized.
The general formula of seating nn people around a circular table should be remembered which is (n−1)!(n - 1)!
Also, the formula for nn people seating in a row is the total number of permutations in it which is n!n!.