combinatorics
How many ways seven people can sit around a circular table?
For first, I thought it was $7!$ (the number of ways of sitting in seven chairs), but the answer is $(7-1)!$.
I don't understand how sitting around a circular table and sitting in seven chairs are different. Could somebody explain it please?
I would do the problem exactly the way you did it.
Here is an alternative approach to confirm your answer.
We can line up the five students in $5!$ ways, leaving spaces between them and at the ends of the row in which to insert the teachers. There are six such spaces, four between successive students and two at the ends of the row. We can insert the three teachers in $P(6, 3) = 6 \cdot 5 \cdot 4$ ways. This gives us $$5! \cdot 6 \cdot 5 \cdot 4$$ linear arrangements of students and teachers in which no two teachers are consecutive.
However, since we wish to arrange the students and teachers around a circular table so that no two teachers sit in consecutive seats, we must exclude those linear arrangements in which teachers are at both ends of the row. There are three ways to select the teacher at the left end of the row, two ways to select the teacher at the right end of the row, and four ways to place the remaining teacher in one of the four spaces between successive students. Hence, there are $$3 \cdot 2 \cdot 4 \cdot 5!$$ linear arrangements in which teachers are at both ends of the row.
Hence, there are $$6 \cdot 5 \cdot 4 \cdot 5! - 3 \cdot 2 \cdot 4 \cdot 5! = (120 - 24)5! = 96 \cdot 5!$$ linear arrangements of teachers and students so that no two teachers are consecutive and teachers are not at both ends of the row.
These linear arrangements correspond to the permissible ways we can seat the students and teachers around the table. To account for rotational invariance, we divide the number of linear arrangements by $8$, which yields $$\frac{96 \cdot 5!}{8} = 12 \cdot 5! = 1440$$ permissible seating arrangements around a circular table, as you found.
We initially seat A, B, and D. Since A and B must sit together, there are only two ways to do this, depending on whether B sits to the left or right of A, as shown below.
To ensure that C and E do not sit together, we must place them to the left or right of the block. There are two ways to do this. Hence, there are four admissible arrangements, as shown below.
Best Answer
In a circular arrangement we first have to fix the position for the first person, which can be performed in only one way (since every position is considered same if no one is already sitting on any of the seats), also, because there are no mark on positions.
Now, we can also assume that remaining persons are to be seated in a line, because there is a fixed starting and ending point i.e. to the left or right of the first person.
Once we have fixed the position for the first person we can now arrange the remaining $(7-1)$ persons in $(7-1)!= 6!$ ways.