# Combinatorics : Circular Permutation of N people with N-1 spots

This question has two parts.

1) Number of ways for $N$ people to sit around a circular table with $N-1$ seats

Firstly, for $N$ people to sit around a circular table with $N$ spots, it is well known that the number of possible permutations is $(N-1)!$.

Now, if there are only $N-1$ spots around the table, I thought that the total number of possible permutations is $N \cdot(N-2)!$.

Here is my reasoning and I wanted to verify it so please correct me if you spot any errors.

Simply, there are ${N \choose N-1}$ ways to choose $N-1$ people out of $N$ people. For each combination thus formed, it is simply sitting them around in a circle so $(N-2)!$. Thus total is ${N \choose N-1} \cdot (N-2)!$

In general, to sit $N$ people around a circular table with $K$ seats, the formula is ${N \choose K} \cdot (K-1)!$

2) Number of ways to mark the numbers 1 ~ 6 on the surfaces of cube (regular hexadron)

Again, here is my reasoning and I wanted to verify my understanding and solution so please correct me if you spot any errors

The first number we pick can go on any of the 6 surfaces and whichever surface we choose to mark on does not matter since we can always put that surface on the bottom (just like how we can start the same permutation on any seat on the circular table). So the number of ways is 1 (since we end up dividing by 6 anyways to get rid of all the duplicates).

So with the first number marked on the bottom surface a, we have five remaining numbers and 5 surfaces left.

Now, we can put any one of the remaining 5 numbers on the surface d, but now we see that the four lateral surface $c, d, e, f$ is really a circular permutation problem: since whichever surface we choose to mark the second number, the bottom is always surface a and the top is f.

i.e. We are trying to permutate the remaining 5 numbers on the 4 lateral surfaces that are circular, so we get ${5 \choose 4} \cdot (4-1)! = 30$ possible ways.

We have only 1 remaining number after this and only surface $b$ left to mark. So again, like the first number, the number of ways is 1.

Total, we have $1 \cdot 30 \cdot 1 = 30$ total ways to mark the numbers 1 ~ 6 on a cube.

Please correct me if you spot any errors!!

• I assume you meant the top surface is $\boldsymbol{b}$. Otherwise, your arguments look good. – N. F. Taussig Mar 10 '18 at 20:06