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Forum index » Science and Technology » Math » Combinatorics
simultaneous pairings
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Rich
science forum beginner


Joined: 22 Aug 2005
Posts: 28

PostPosted: Thu Oct 13, 2005 11:54 pm    Post subject: simultaneous pairings Reply with quote

My wife, a high school French teacher, has 5 classes of 20 to 30 students
each. She often has the students work in pairs and wants to create the
maximum number of lists of unique simultaneous pairings of students for
each
class. I boldly stated that it was a task custom-made for a computer and
that I would implement it.

The number of possible pairings is n!/((n-2)!2!), and, using Visual Basic,
my routine
produces all the possible pairs just fine.

The kicker is the creation of the unique lists. The problem arises because
of the need for the pairs to work simultaneously, and there can obviously be
no duplication of one student in any given list.

I've searched in vain for a pattern of pair selection that leads to the
automation of the process of producing the lists.

Any ideas?

Thanks.

Rich
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Proginoskes
science forum Guru


Joined: 29 Apr 2005
Posts: 2593

PostPosted: Sat Oct 15, 2005 8:03 am    Post subject: Re: simultaneous pairings Reply with quote

Rich wrote:
Quote:
My wife, a high school French teacher, has 5 classes of 20 to 30 students
each. She often has the students work in pairs and wants to create the
maximum number of lists of unique simultaneous pairings of students for
each
class. I boldly stated that it was a task custom-made for a computer and
that I would implement it.

The number of possible pairings is n!/((n-2)!2!), and, using Visual Basic,
my routine
produces all the possible pairs just fine.

The kicker is the creation of the unique lists. The problem arises because
of the need for the pairs to work simultaneously, and there can obviously be
no duplication of one student in any given list.

I've searched in vain for a pattern of pair selection that leads to the
automation of the process of producing the lists.

Any ideas?

Thanks.

Google for "Whist tournaments". This will give solutions for groups of
3 or 4 people, so the method should apply, or you will find out it's
easy because _____ . (I don't know how to fill in the blank here.)

--- Christopher Heckman
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Google

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