How many different arrangements are possible? 12 students, no single table for 12. They need to sit around 2 circ tables with 6 chairs.

Question

Twelve students are going for lunch . There is no single table for twelve. So, they will sit around two circular tables, each with six chairs. How many different arrangements are possible? (original question)

From what I understand in this question, 12 students in two circular tables with 6 chairs each. I have an answer but I'm not sure if I'm correct or wrong. 12C6(number of chairs) = 924, so I've multiplied it by two because there's six chairs in two circular table. Getting the answer 1848 different arrangements. I just want to know if I did it right. If I'm wrong in what ways can I answer the question? Thanks

Answer 1

First of all, does the order of the tables matter? The question doesn't explicitly state this (at least the way I read it), so we'll have to make an assumption about this.

Your answer has a few problems. Finding $12C6$ is good because it tells you how to find $6$ people sitting at one table- furthermore, it also decides the $6$ people sitting at the other table. I'm not sure why you multiplied by $2$; if the order of the tables does matter, then $12C6$ is fine as is. If the order does not matter, then you must divide by $2$.

Now that you have $6$ people at one table, and $6$ people at the other table, in how many ways can you arrange $6$ people at a table? This has to do with circular permutations, and can be answered simply: if you're trying to arrange $6$ people at a table in a circle, all of the seats are considered indistinct at first, because the seats don't matter, but the position of the people relative to each other. Once you've seated the first person, every other seat is determined by position relative to the first person, so each of the seats is now distinct. That means there are $5!$ ways to seat the remaining $5$ people. Since you have to do this twice, there are two factors of $5!$ you need to take into account.

Answer 2

If the question means to count the number of ways the students may be assigned to the two tables, without regard for the ordering within each table, then your answer is off by that factor of two. After choosing which six sit at table $A$, there is no choice about which six sit at table $B$.

However, you really need to know what the question is asking. Probably there is interest in the order of seating at each table. Very probably, because the table is circular, seating in the order 1,2,3,4,5,6 is the same as the order 6,1,2,3,4,5. But does it matter whether students 1,2,3,4,5,6 are at table $A$ or $B$? And so forth.

Answer 3

Several interpretations could be put on the question, but I am specifying one set of assumptions, to focus on the multiplier of $2$, on which some wrong/vague views have been expressed.

• The tables are labelled

• The seats are unnumbered, so rotating an arrangement won't yield a new one, and the formula $(n!/n) = (n-1)!$ will apply

Under these assumptions, the # of arrangements $= \binom{12}6\cdot (5!)^2$

There will not be any multiplier of $2$, even though the tables are labelled.
To understand why, consider seating two people $(A,B)$ at two tables $(1,2)$:
$\binom21 = 2\;$ yields $A1-B2\;and\; A2-B1,\;\;$ i.e. covers all possibilities !

In fact, it is the other way round. Had the tables been unlabelled, we need to divide by $2$.

Answer 4

There are ${12\choose 6}$ ways of separating the $12$ students into $2$ groups of $6$. There are then $5!$ ways of permuting $6$ students in a circle. Hence the total number of seating arrangements is

$$ {12 \choose 6} (5!)^2.$$

This assumes that the tables are labelled i.e if all the students got out of their chairs and swapped to the other table, then this would be a different arrangement. If you don't want to count these as separate arrangements, then you can divide the above by $2$.

Hope this helps.