# Combinatorics in a Party.

There are 12 persons in a dinner party, they are to be arranged on two sides of a rectangular table. Supposing that the master and the mistress of the house have are always facing each other, and there are two specific guest who must always, be placed alongside one another. Find the number of ways in which the company can be placed.

Number of ways master and the mistress of the house can be seated is $2 \cdot$$5 \choose 1$$=10$

Now, one position on every side is now fixed, Let us consider two specific guests as one element with two internal arrangements. For any side, we have to choose one position out of 3 but due to internal arrangements and two sides their arrangement becomes $2\cdot 2\cdot 3=24$

For remaining 8 persons the number of arrangements is $8!$.

Total Possible Arrangements = $10 \cdot 24 \cdot 8!$.

• Is this a question? – amcalde Dec 15 '14 at 15:18
• If master and mistress have fixed seats, then there is only one way for them to be seated, not $50$. – drhab Dec 15 '14 at 15:25
• no, they only have to opposite to one another – Dheeraj Kumar Dec 15 '14 at 15:25
• Then that should have been said. Not that they have fixed seats. – drhab Dec 15 '14 at 15:26
• "opposite to one another" – Dheeraj Kumar Dec 15 '14 at 15:27

I now understand the problem in the following way: There is a table with dimensions $6\times 1$ in which for some reason people don't sit on the short edge. We wan't to count the number of ways to seat them so that the hosts are sitting in front of each other and the members of the couple are seated next to each other. We seperate in three cases:

Case 1: the couple hosts sit at the spots at the corner. There are two ways to select the side of the rectange and there are two ways to select which host sits at which of the two seats. After this notice the couple must sit at consecutive seats, how many pairs of consecutive seats are there? four on each side of the table, so eight. After this we must select which member of the couple sits in which of the two spots in two ways, and finally seat everyone else in $8!$ ways, so the answer is $2\cdot2\cdot8\cdot2\cdot8!$

Case 2: the hots sit at spots that are neither at the middle of the table or at the edges, there are $2$ ways to select the pair of spots they want, after this $2$ ways to select which couple takes which of the two seats. We now sit the couple, there will be three pairs of consecutive seats on each edge so there are $6$ ways to chose the pairs, $2$ ways to select which member of the couple takes which seat and $8!$ ways to sit everyone else. The answer is thus $2\cdot2\cdot6\cdot2\cdot8!$

Case 3: the hosts sit at central spots,there are two ways to select which of the central spots, then there are two ways to select which host gets which of the two spots, notice there are three pairs of consecutive spots left on each side, so there are $6$ ways to select the spots the couple sits at, and then $2$ ways to select which member of the couple gets which seat, after this $8!$ ways to sit everyone else, hence there are $2\cdot2\cdot6\cdot2\cdot8!$

factoring the $8!$ the final answer is:

$(64+48+48)8!=160(8!)$

• Case 3 is wrong: there are two "middle" seats on each side, and two ways the hosts can sit in the selected one. This leaves six pairs of places for the adjacent couple (three on each side; one pair to one side of the hosts and two to the other) again times 2 for which sits where, so the last term should be 2.2.6.2.8! for a total of 160.8! – TripeHound Dec 15 '14 at 16:30

A simpler way of looking at it is to consider the couple who must be adjacent first. There are five adjacent pairs of seats on each side of the table, and the couple can sit either way around (5 * 2 * 2). Wherever they sit, there are four places left where the hosts can sit opposite each other, and either host can sit either side (...* 4 * 2). The remaining 8 guests can sit 8! ways. Thus the total is 5 * 2 * 2 * 4 * 2 * 8! or 160 * 8!

• Good idea, I don't know why I jumped to classify according to host instead of couple. – Jorge Fernández Hidalgo Dec 15 '14 at 16:36
• I started to think that way as well, presumably because it's mentioned first in the question, but realised the adjacent pair is the more limiting. – TripeHound Dec 15 '14 at 16:38