# How many ways to divide 13 people into groups of 2 without repeating combinations?

I have $$13$$ people and need them broken into groups of $$2$$ where everyone gets to meet with each other at least $$1$$ time.

If each person is assigned a letter ($$A$$, $$B$$, $$C$$, etc) how can I organize a chart to get this list started?

• There is no way to divide an odd number of people into groups of two... – 5xum Jun 10 at 13:21
• I reckon the number of groups of 2 that have to form is the 12th triangle number 1+2+3+4.....+11+12 A has to meet 12 people, B, 11 . By the time we get to the 13th, m or whatever, he has no one left he needs to meet. My answer 78 – Cato Jun 10 at 13:27
• This question is unclear to me. It sounds as though you are asking for an example of sequence of "rounds" in which you break the people up into pairs with an odd person left out such that by the end each person has been paired with each other person at least once. Or, are you actually asking for the number of ways in which this can happen? – JMoravitz Jun 10 at 13:37

I believe you mean how can they all meet with each other once as you cannot divide 13 by 2.

If we let the group be of Size 3.

We have A, B,C

A seeing B is denoted A -> B

We want:

A-> B

A-> C

B -> C

If we add another participant D, then we want

A -> D

B -> D

C -> D

Note that since A has seen 3/3 people, then B only needs to see the other two C and D, then C only needs to see D. The next person only needs to see n-1 the previous person.

We can generalise with:

$$(n-1)!$$

So your chart will have $$12!$$ pairings

Imagine seating them at a table with seven people on one side and six plus an empty chair on the other. Each meets with the person across the table. After one round, everybody moves one seat clockwise. The empty seat stays empty. Each person meets with the person opposite again. After $$13$$ rounds each person will have met everybody else and sat out for one round. There are six meetings per round for a total of $$78$$, which is the number of pairs of people.