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

permutations &combinations

### 2 Answers by Expert Tutors

Richard P. | Fairfax County Tutor for HS Math and ScienceFairfax County Tutor for HS Math and Sci...
4.9 4.9 (638 lesson ratings) (638)
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

Few people take into account the rotational symmetries.  Nice application of forced exclusion in your explanation.
Vivian L. | Microsoft Word/Excel/Outlook, essay composition, math; I LOVE TO TEACHMicrosoft Word/Excel/Outlook, essay comp...
3.0 3.0 (1 lesson ratings) (1)
-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

You have neglected to account for the rotational symmetries.  To see why, consider the case of 3 people at a single table.  Your solution would indicate 3!=6 possible arrangements, namely (read the arrangement clockwise)

ABC
ACB
BAC
BCA
CAB
CBA

Note, however, that there are exactly two equivalence classes of arrangement that group as

ABC, BCA, CAB
ACB, BAC, CBA

Hence there are exactly two ways for three individuals to sit around a round table:

ABC, ACB.

In fact, given N people with N seats at a round table, there are N!/N = (N-1)! ways to arrange the individuals.  As Richard alludes to in his solution, it is a matter of wording for the equivalency of seating arrangements.  Oftentimes the implied solution for these types of problems is that two seating arrangements are considered equivalent if the person to the right of each individual does not change.  (See my example above for a better understanding.)