# Thread: identical round tables

1. ## identical round tables

There are 2 identical round tables.At least 1 person must sit around
every table. In how many ways 5 persons may sit?

This is the solution:
C(5,1).0!.3!+C(5,2).1!.2!

How do we get 0! and 1!?

2. ## Re: identical round tables

The number of ways that N people can sit at a round table is (N-1)!, assuming that the only important things is the relatie position of each person. So for example for 3 people there are only 2 arrangements: either B sits to A's left and C to A's right, or vice vesa. One way to think of it us that the first person can sit anywhere - it doesn't matter in which sea - then the 2nd person can sit in any one of (N-1) remaining seats, the third in (N-2) remaining seats, etc.

The solution as shown considers that there are two possible cases: 1 person sits at one table and 4 at the other, or 2 people sit at one table and 3 at the other. It does not matter which table is which, so if person A sits at table 1 and persons B, C, D and E at table 2 it's the same as person A sitting at table 2 and persons B, C, D and E sitting at table 1. For the case of 1 person at one of the tables there are C(5,1) was to choose who that 1 person is, and (N-1)! = 0! ways to arrange that one person at his table. Then for the remaining 4 people there are (N-1)!=3! ways to arrange them at the other table. For the case of 2 people at one of the tables there are C(5,2) ways to pick those two people, and (N-1) = 1! ways to arrange them at that table, and then 2! ways to arrange the other three at the other table. Hope this helps!

many thanks.

4. ## Re: identical round tables

Hello, kastamonu!

ebaines is correct.
Let me give it a try . . .

There are 2 identical round tables. At least 1 person must sit at each table.
In how many ways 5 persons may sit?

This is the solution:
$C(5,1)\!\cdot\!0!\!\cdot\!3! + C(5,2)\!\cdot\!1!\!\cdot\!2!$

How do we get 0! and 1!?

As he noted: $n$ people can sit around a round table in $(n-1)!$ ways.

There are two cases to consider.

[1] One at one table and Four at the other.
. . .There are $C(5,1)$ ways to choose the One person.
. . .He can sit at his table in $0! = 1$ way.
. . .The other Four people can sit at their table in $3!$ ways.
Hence: . $C(5,1)\!\cdot\!0!\!\cdot\!3!\text{ ways.}$

[2] Two at one table and Three at the other.
. . .There are C(5,2) ways to choose the Two people.
. . .They can sit at their table in $1!$ way.
. . .The other Three people can sit at their table in $2!$ ways.
Hence: . $C(5,2)\!\cdot\!1!\!\cdot\!2!\text{ ways.}$

Therefore: . $C(5,1)\!\cdot\!0!\cdot\!3! + C(5,2)\!\cdot\!1!\!\cdot\!2!\text{ ways.}$

Many Thanks