Question

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

Answer 1

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

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

Complete step-by-step solution:
We know that there are a total $2n$ numbers of people which have to be seated.
There have to be $n$ number of people selected out of the total $2n$ number of people, this can be found using: ${}^{2n}{C_n}$
This can be simplified by using the formula as: $(2n!)(2n - n)!$
On subtracting the bracket terms we get, $(2n!)(n)!$
Now for the first $n$ which have to be seated at a round table:
The first person out of the $n$ people will have total $n$ 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 $n$ people can be seated round a table is:
$(n - 1) \times (n - 2).... \times 1$
This could be also represented as $(n - 1)!$.
Now the total numbers of ways $n$ items can be arranged are in $n!$ 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 $2n$ number of people can be seated will be:
Selection of \[n\] people out of ${{2n}} \times {\text{n}}$ people sitting in a circle ${{ \times n}}$ people sitting in a row
This could be written as:
=$(2n!) \times n! \times (n - 1)! \times n!$
On multiplying both we get:
=$2n!{(n!)^2}(n - 1)!$

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

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