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!?
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!
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:
$\displaystyle C(5,1)\!\cdot\!0!\!\cdot\!3! + C(5,2)\!\cdot\!1!\!\cdot\!2!$
How do we get 0! and 1!?
As he noted: $\displaystyle n$ people can sit around a round table in $\displaystyle (n-1)!$ ways.
There are two cases to consider.
[1] One at one table and Four at the other.
. . .There are $\displaystyle C(5,1)$ ways to choose the One person.
. . .He can sit at his table in $\displaystyle 0! = 1$ way.
. . .The other Four people can sit at their table in $\displaystyle 3!$ ways.
Hence: .$\displaystyle C(5,1)\!\cdot\!0!\!\cdot\!3!\text{ ways.}$
[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 $\displaystyle 1!$ way.
. . .The other Three people can sit at their table in $\displaystyle 2!$ ways.
Hence: .$\displaystyle C(5,2)\!\cdot\!1!\!\cdot\!2!\text{ ways.}$
Therefore: .$\displaystyle C(5,1)\!\cdot\!0!\cdot\!3! + C(5,2)\!\cdot\!1!\!\cdot\!2!\text{ ways.}$