There are two dinstict round tables,each with 5 seats,in how many ways may a group of 10 be seated

Question

permutations &combinations

Answer 1

If the seating order around each table does not matter, this problem boils down to the number of ways that 5 people can be chosen from a group of 10 people.   The short way of saying this is: "ten choose five". 

The logic here is that once the first table is selected, the remaining people must go to the second table.

 

This number is the combinatorial coefficient   10!/[ 5! (10-5)!]            (x! denotes  x factorial)

 

This evaluates to 252

 

If the seating order around a table does matter this number must be multiplied by the number of circular permutations of 5.    This number is 4! = 24.  In fact, one must multiply by 24 twice, once for each table.  

 

So if the seating order does matter, then the answer is 252 x 24 x24 = 145152

 

 

 

Answer 2

Hi Joshua;

The fact that there are two tables is a non-issue.  What if there is only one table?  There are ten seats, either way.

For the first seat, there are 10 possibilities.

For the second seat, there are 9 possibilities.

For the third seat, there are 8 possibilities.

For the fourth seat, there are 7 possibilities.

For the sixth seat, there are 6 possibilities.

etc.

10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1=3,628,800