Question

How many ways are there of seating `10` couples around a round table such that each couple is sat together?

Answer 1

`2*10!*2^10 = 7431782400`

Explanation:

I will assume that there are `20` fixed seating positions around a table such that the 'last' place is adjacent to the 'first'. Let's label the positions `1,2,...,20`.

The couples are either arranged in positions:

`1&2, 3&4, 5&6,..., 19&20`

or:

`20&1, 2&3, 4&5,..., 18&19`

Each of these possibilities results in `10` "couple positions", resulting in a total of `10!` possible orderings of couples.

So we have a total of `2*10!` possible orders of couples.

Then whichever of these orders is chosen, there are two possible orders within each couple, resulting in `2^10` possibilities.

That gives a total of:

`2*10!*2^10 = 2*3628800*1024 = 7431782400`

arrangements.