Okay...

According to Pete's method, I come up with 15 unique games possible.

However, utilizing Bill's method, I get 20 for the permutations and 10 for
the combinations.

Is there a formula or function that will actually figure this correctly?

Thanx.

"Pete_UK" wrote:

If it was just one player against another, you can set this up in a
simple table like this:

P B R O G
Purple (P) x y y y y
Blue (B) y x y y y
Red (R) y y x y y
Orange (O) y y y x y
Green (G) y y y y x

the x's in the table indicate that P cannot play P etc. The table
covers home and away type fixtures, with the home side in column A and
the away side in row 1. If this is a golfing competition, for example,
where this does not apply, then you can remove the triangle of matches
above the leading diagonal of x's, so tht you have something like this:

P B R O G
Purple (P) x
Blue (B) y x
Red (R) y y x
Orange (O) y y y x
Green (G) y y y y x

You can apply the same principle with pairs - you just need to write
down all possible combinations. Assuming P & B is the same as B & P,
you would have this down column A:

PB
PR
PO
PG
BR
BO
BG
RO
RG
OG

Repeat the list going across, ignore the diagonal, and count the y's.

Is this what you wanted? You can do this on a piece of paper or type it
all into Excel and then print it out onto a piece of paper!

Hope this helps.

Pete


Steven Sinclair wrote:
Here's what I have...

Purple
Blue
Red
Orange
Green

Each of the colors represent a player.
Each of the players needs a partner.
Each set of partners needs to play with another set of partners.
Each player needs to play with a different partner against all other partners.

All-in-all, we need to count how many games can be played with how many
UNIQUE partners and UNIQUE opposing partners.

I have no idea of even where to begin other than just writing everything out
by hand on a piece of paper.

Thanx.