# There are two distinct round tables, each with 5 seats. In how many ways may a group of 10 be seated?

6048

#### Explanation:

There's a few ways to approach this - let me show you one.

Let's first notice that when dealing with seating at tables, we'll have to work with the lack of definite seat numbers (as opposed to a row of seats that has definite end seats). So what I want to do is break down the problem into the numbers of ways I can have 10 people sit at the two tables, then deal with actual seating.

Putting people at the tables

If we take Table 1 and choose 5 people to be at it, we'll naturally deal with Table 2 (if the people aren't at Table 1, they're at Table 2).

So let's take 10 choose 5:

C_(n,k)=((n),(k))=(n!)/((k!)(n-k)!) with $n = \text{population", k="picks}$

$\left(\begin{matrix}10 \\ 5\end{matrix}\right) = 252$

One thing to notice - when I have persons 1, 2, 3, 4, 5 at Table 1 and 6, 7, 8, 9, 10 at Table 2, it's the same as having persons 6, 7, 8, 9, 10 at Table 1. So to get rid of the duplicates, let's divide by 2:

$\frac{252}{2} = 126$

Seating at the tables

For each table, we have 5 people sitting. If we were dealing with rows of chairs, we'd have 5! =120 ways to seat the people. But since these are round tables, we have to divide through by the number of seats to get rid of the duplicates (seating 1, 2, 3, 4, 5 is the same as seating 2, 3, 4, 5, 1), which leaves 4! =24 per table.

There are two tables, so we multiply by 2:

$24 \times 2 = 48$

Putting it together

There are 126 ways to divide the people up into distinct groups at the tables and 48 ways to seat people per group, giving:

$126 \times 48 = 6048$ different ways to seat people.