# [HELP] Calculating all colour-balanced combinations of two-colour factions

I need a little bit of help with math x)

I'm trying to find all the possible combinations of five pairs of two colours (among the five MTG colours) where each colour appear exactly twice. You all know the classic two examples of all allies and all ennemies colours:

ALLIES: WU UB BR RG GW

ENEMIES: WB UR BG RW UG

But there are many others, for instance the ones they used in the most recent Ravnica sets:

GUILDS OF RAVNICA: UB UR BG RW GW

RAVNICA ALLEGIANCE: WU WB BR RG GU

I took a pen and paper and tried to find them all in a logical order. Now, if I didn't miss something, I believe there are 12 possible combinations:

01: WU UB BR RG GW (Allies)

02: WU UB BG RG RW

03: WU GU RG BR WB (Ravnica Allegiance)

04: WU GU BG BR RW

05: WU UR BR BG GW

06: WU UR RG GB WB

07: WB UR BG RW GW (Enemies)

08: WB UB GU RG RW

09: WB UB UR RG GW

10: WB BR UR GU GW

11: WR BR UB GU GW

12: WR UR UB BG GW (Guilds of Ravnica)

So, here's my question: Does someone know where to find the exact answer to this question?

Bonus question: Does someone know how to calculate mathematically the number of possibilities to make sure I didn't miss anything?

Thanks in advance to whoever can help

Have a good day =D

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## Comments

Awesome thanks! (Out of curiosity, do you know how he calculated it?)

(Sorry, got a bit excited here.)

Maybe I can give it a shot...

We can ensure we get a color balanced arrangement as follows:

1. Arrange the 5 colors in a line.

2. Read the guilds off the line.

E.g. If we have WBGUR, then we have WB, BG, UG, UR, and (wrapping around) WR.

There are 5!=120 ways to arrange the 5 colors, but note that sometimes it doesn't matter which color you start with. e.g. WUBRG and UBRGW give the same guilds. For each arrangement, there are 4 similar other arrangements that produce the same guilds, so we need to divide by 5.

Moreover, we can have the same guilds but with the other colors going the other direction e.g. WUBRG and WGRBU. So we need to divide by 2.

So we have a total of 5!/5/2 = 12 color balanced arrangements.

Would love to see a cleaner (and more rigorous!) explanation

Ooooh that's genius, you just have to think of the line thing x) Thanks a lot! =D