Math Puzzle? Help?

Being bad at math means I’m not even good at describing what my problem is here, so anyway let’s give this a shot.

I am trying to work out how much work is involved in writing text for a number of combinations of cards. What we start with are two sets of cards to start with – 1, 2, 3, and A, B, C. At the start of the game, a player removes one card from each set (at random, but that’s not important). Then I want to work out how many permutations there are of the remaining sets.

This isn’t as easy as ‘6 choose 4’ – that’s where I started. Because there are some permutations that are impossible. You can’t possibly have ABC1 or 123A. There needs to be one missing from each set of cards.

So.

Anyone know how to work out what I’m talking about? Is this something where it’s easier to run a small script or something?

1 Comment

  1. I’m not sure if you monitor the comments, or if I’ve understood the problem correctly, but here goes:

    You’ve got three cards per set, so there’s three different ways you can remove a
    card from a set. This gives you three possible subsets of remaining cards for
    each of 123 and ABC:


    23 13 12
    BC AC AB

    These can be combined in 3^2 = 9 different ways:


    23BC 23AC 23AB
    13BC 13AC 13AB
    12BC 12AC 12AB

    Each such set can be arranged in 4! = 24 different ways. 9 possible sets of
    remaining cards, 24 possible arrangements each gives 9 * 24 = 216.

    For completeness sake, here are those arrangements (I used a Python script here,
    for obvious reasons):


    AB12 AB21 A1B2 A12B A2B1 A21B BA12 BA21 B1A2 B12A B2A1 B21A
    1AB2 1A2B 1BA2 1B2A 12AB 12BA 2AB1 2A1B 2BA1 2B1A 21AB 21BA
    AB13 AB31 A1B3 A13B A3B1 A31B BA13 BA31 B1A3 B13A B3A1 B31A
    1AB3 1A3B 1BA3 1B3A 13AB 13BA 3AB1 3A1B 3BA1 3B1A 31AB 31BA
    AB23 AB32 A2B3 A23B A3B2 A32B BA23 BA32 B2A3 B23A B3A2 B32A
    2AB3 2A3B 2BA3 2B3A 23AB 23BA 3AB2 3A2B 3BA2 3B2A 32AB 32BA
    AC12 AC21 A1C2 A12C A2C1 A21C CA12 CA21 C1A2 C12A C2A1 C21A
    1AC2 1A2C 1CA2 1C2A 12AC 12CA 2AC1 2A1C 2CA1 2C1A 21AC 21CA
    AC13 AC31 A1C3 A13C A3C1 A31C CA13 CA31 C1A3 C13A C3A1 C31A
    1AC3 1A3C 1CA3 1C3A 13AC 13CA 3AC1 3A1C 3CA1 3C1A 31AC 31CA
    AC23 AC32 A2C3 A23C A3C2 A32C CA23 CA32 C2A3 C23A C3A2 C32A
    2AC3 2A3C 2CA3 2C3A 23AC 23CA 3AC2 3A2C 3CA2 3C2A 32AC 32CA
    BC12 BC21 B1C2 B12C B2C1 B21C CB12 CB21 C1B2 C12B C2B1 C21B
    1BC2 1B2C 1CB2 1C2B 12BC 12CB 2BC1 2B1C 2CB1 2C1B 21BC 21CB
    BC13 BC31 B1C3 B13C B3C1 B31C CB13 CB31 C1B3 C13B C3B1 C31B
    1BC3 1B3C 1CB3 1C3B 13BC 13CB 3BC1 3B1C 3CB1 3C1B 31BC 31CB
    BC23 BC32 B2C3 B23C B3C2 B32C CB23 CB32 C2B3 C23B C3B2 C32B
    2BC3 2B3C 2CB3 2C3B 23BC 23CB 3BC2 3B2C 3CB2 3C2B 32BC 32CB

    I suppose in general, if you have X sets of Y cards each, then the number of
    possible arrangements after removing one card from each set is:


    (X * (Y - 1))! * Y^X

    You should run the numbers through someone else though, because I also suck at
    math, and I’m pretty sure I messed this up somewhere :p

    Good luck with your thing! :3

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