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How many combinations?


Mortimer
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Spoiler

8*6!

extension: 64*6!

Reasoning:

Spoiler

The layout doesn't matter except to know that there are 6 positions/cards.

If there were X unique cards (and ignoring rotations), there are X! possible permutations.

If there are X cards of type 1, Y cards of type 2, and Z cards of type 3, and again ignoring rotation, you could arrange them in (X+Y+Z)!/(X!*Y!*Z!) ways.**

It is easy to see how this can be extended to any number of groups of any size.

Now to deal with the rotations.  Each position can be right-side-up or up-side-down.  So for each of the ways you can rearrange their positions, there are 2^6 ways the cards could be rotated.

So the answer to the original problem is (2^6)*(2+2+2)!/(2!*2!*2!) = (2^6)*6!/2^3 = 8*6!

The extension is (2^6)*(1+1+1+1+1+1)!/(1!*1!*1!*1!*1!*1!) = 64*6!

One thing to note is the formula for number of combinations is a special case of the above formula(**).  So if there are k of the type you are interested in, and n-k others, the number of combinations of n taken k at a time (aka n choose k, or nCk) is (k+(n-k))!/(k!*(n-k)!) = n!/(k!*(n-k)!)

 

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