Unique Combinations

jsmith13 · 2007-01-15 16:03:05

This is a real life problem. I am organizing a networking meeting

85 people will attend
We will seat them at 17 tables with 5 people at each table
At the first seating (while the soup is served) everyone will introduce themselves and present the group with a question or problem they have prepared

After the soup and discussion, everyone will be sent to new tables where salad will be served and each person will again introduce themselves and persent their question

After salad, everyone will again be sent to new tables where the entre will be served and again they will present their question

Finally, everyone will go to a new table for desert and their introduction and question

We want to make sure that no one sits with anyone they have been seated with during the earlier seatings

Is there a formula or system I can use to assign seating so that everyone gets to meet 16 different people (4 seatings with 4 new people each) with no repeats?

Trial and error is just giving me a headache!

Can I use the same formula or system if 90 people show up and we put them at 18 tables of 5?

What if we change our minds and go with tables of 6?

Thanks for any help I can get

JimS

pi man · 2007-01-15 17:33:19

I'm not positive this would work but it might.

Each seat is assigned a table letter (A-Q) and table position (1-5). So there are five seats at table A: A1, A2, A3, A4, A5.

Assign every one a seat for the first session. For each subsequent session, each person willl always be seated in the same position at a table as the position they were for the first session. Different table probably, but same position.

People in position 1 at each table stay where they're at. So the person in seat A1 will be there for every session.

The people in position 2 at each table go to the next lettered table and sit in the same position as they were at their first table. The person in A2 would go to B2, then C2 and then D2.

The people in position 3 would skip over a table and go to the next one: A3, C3, E3, F3
The people in position 4 would skip over 2 tables and go to the 3rd table: A4, D4, G4, J4
The people in position 5 would skip over 3 table and go to the 4th table: A5, E5, I5, M5

Consider the tables to be in a circular pattern. If you trying to move past table Q, start over at table A. So the person starting at N5 would go to A5, E5 and then I5.

You need to map this completely out before you implement to make sure it works. I got as far as scheduling half of the people and didn't run into any problems.

MathsIsFun · 2007-01-15 18:34:16

Sounds very workable. Position 5 will be at table 17 after 4 moves. Well done, pi man!

Using your idea, this is how 8 tables would go for 10 moves (not that you would have 10 courses!):

(1,1,1,1,1,1,1,1)
(1,2,3,4,5,6,7,8)
(1,3,5,7,9,11,13,15)
(1,4,7,10,13,16,2,5)
(1,5,9,13,17,4,8,12)
(1,6,11,16,4,9,14,2)
(1,7,13,2,8,14,3,9)
(1,8,15,5,12,2,9,16)
(1,9,17,8,16,7,15,6)
(1,10,2,11,3,12,4,13)
(1,11,4,14,7,17,10,3)

Note: this is only looking at where the people at table 1 are seated. In other words, they are all at table 1 to start with (1,1,1,..), then the first one is still at table 1, the second at table 2, etc, or in short hand (1,2,3,...), etc.

For people at table 2, just add 1 to every number, so you would have (2,2,2,...) then (2,3,4,...), etc

(Once you get all that sorted out, you could add one minor enhancement: add 1 table to every move, so every one gets to move. You could even label the tables randomly around the room, so that the person who moves 1 table gets to move around the room a lot. You could also add a position rotation to everybody, too.)

Interestingly, with only 16 tables, it breaks down pretty quick:

(1,1,1,1,1,1,1,1)
(1,2,3,4,5,6,7,8)
(1,3,5,7,9,11,13,15)
(1,4,7,10,13,16,3,6)
(1,5,9,13,1,5,9,13)

So, 17 being prime is probably a big help.

18 tables works like this:

(1,1,1,1,1,1,1,1)
(1,2,3,4,5,6,7,8)
(1,3,5,7,9,11,13,15)
(1,4,7,10,13,16,1,4)
(1,5,9,13,17,3,7,11)
(1,6,11,16,3,8,13,18)
(1,7,13,1,7,13,1,7)

(Don't forget the "Seating Arrangements Courtesy of Math Is Fun Forum" ... just kidding!)

pi man · 2007-01-16 02:51:26

Yes, 17 being prime is a big part of it. Because it is prime, people in positions 2-5 will never sit at the same table twice (until you have more than 17 sessions). That's critical because those in position 1 don't move.

So Jim, you can either figure out how to handle eighteen tables or limit it to 17 and turn it into a marketing ploy - "Due to popular demand, seating is unlimited and must be reserved in advance. First come, first served. Get your seat TODAY!"

mathsyperson · 2007-01-16 03:23:47

pi man wrote:

"Due to popular demand, seating is unlimited and must be reserved in advance."

Hee hee...

I can't laugh at you too much though, that was a very clever method you thought up. It's actually quite flexible as well, because although it only works for prime numbers, it works for ALL prime numbers and when the amount of tables is in the tens and twenties then primes are quite common. So, if 10 more people suddenly wanted to come to the meeting, then you'd just grab two more tables and you'd be all set.

In fact, if you were allowed to vary the amount of people on the tables as well, then it work work even better. Let's say you had 91 people who wanted to attend. You'd get 13 tables and allocate 7 people to each of them and everyone would be happy.

So, yes. Very well done.

Ricky · 2007-01-16 04:26:13

If you seat one person at every table, it will also work.

mcassidy · 2013-10-04 07:03:41

I am trying to define the number of unique combinations where: I have three mutually exclusive sets (A,B,C), each of which has a varying number of elements, viz.(A=6; B= 7; and C = 3), and the combinations must have three elements, with one and only one element from each of the three sets. I don't know why I am having difficulties with this, but, alas, I am. Any help would be very much appreciated.
Thanks

bobbym · 2013-10-04 07:15:51

The elements in these sets, are they the same or are they different?

It is usual to speak of set elements as different but there are multisets where elements are the same.

If they are all different there is

6 x 7 x 3 = 126 ways

If they are all the same in each set then there is only 1 way.

mcassidy · 2013-10-04 10:16:29

Each of the elements in each set are different from one another. For example, A1 is different than A2. So also, the elements in set A are different than those in set B, e.g. A1 is different from B3. I am comfortable with the notion of n things taken r at a time, but I frankly never ran into a situation with a constraint such as - given that one may only choose 1 element from set A, etc. The answer you propose, 126, evokes a familiar but long neglected memory.
I very much appreciate your counsel. I have managed to maintain a love of mathematics, despite it remaining a foreign and ineluctable domain.
Thank you!

bobbym · 2013-10-04 11:42:24

Hi mcassidy;

Welcome to the forum. I too have always loved mathematics, it just never loved me.

Math Is Fun Forum

#1 2007-01-15 16:03:05

Unique Combinations

#2 2007-01-15 17:33:19

Re: Unique Combinations

#3 2007-01-15 18:34:16

Re: Unique Combinations

#4 2007-01-16 02:51:26

Re: Unique Combinations

#5 2007-01-16 03:23:47

Re: Unique Combinations

#6 2007-01-16 04:26:13

Re: Unique Combinations

#7 2013-10-04 07:03:41

Re: Unique Combinations

#8 2013-10-04 07:15:51

Re: Unique Combinations

#9 2013-10-04 10:16:29

Re: Unique Combinations

#10 2013-10-04 11:42:24

Re: Unique Combinations

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