How many possibilities are there to place n persons on a round table with $2n$ chairs, so that there is at least one free chair between each $2$ persons?
Assuming we have $n = 10$ persons that want to sit there. So we have $n = 20$ chairs. The first person has $20$ choices. After the first person is placed onto his seat, we have $17$ choices, because the seats next to the first person can't be chosen…
So can I say the number of possibilities is $$\sum_{n=0}^n n*(n-3)$$
Or is that totally false?
And how many possibilities are there to place $n$ persons on a round table with 2n+1 chairs, so that there is at least one free chair between each $2$ persons?
Best Answer
$2n$ chairs.
There are $2$ choices for the chairs that will be possessed by persons.
After that choice there are $n!$ possible arrangements for the persons, so there are $2\cdot n!$ possibilities.
$2n+1$ chairs.
There will be exactly $2$ consecutive chairs that will not be possessed and selecting them comes to the same as selecting the chairs that will be possessed. This gives $2n+1$ choices.
After that choice there are $n!$ possible arrangements for the persons, so there are $(2n+1)\cdot n!$ possibilities.