Charles Ward
Japan Matsumoto Nagano

Here is something I want to learn about  combinations and probability. Sometimes the information on the web a little too advanced, so I will be posting questions and adding solutions to this top post as I go.
CARD COMBINATIONS (choosing one side or the other, some cards optional)
I have 8 cards that are required, but can be placed faceup or facedown. Then there are 6 cards, each of which can be added to the game, faceup or facedown, or not added to the game.
2^8 * 3^6 = 186624
The 2 being the two choices for the 8 cards that must be included. (face up, facedown). And the 3 being the three choices for the cards that may be included (face up, facedown, not included).

Charles Ward
Japan Matsumoto Nagano

I have a set of 4 cards.
Each card has 2 sides. I can choose to have any number of cards. Each card can only have 1 side showing.
1A 1A 2A 1A 2A 3A 1A 2A 3A 4A 1A 2A 3A 4B 1A 2A 3B 1A 2A 3B 4A 1A 2A 3B 4B 1A 2B 1A 2B 3A ...
Would it be better if each card had 3 sides: A, B and N (N meaning Not there)?
How many combinations are possible? 81?
Thanks.

Sam Eckels
United States Wisconsin

It depends on some assumptions, but at first blush I'd say 81 is right and your reframing to A/B/N is the right way to see it.
A few assumptions that didn't come up in your framing:
1A 2A is the same as 2A 1A (i.e. ordering doesn't matter)
1A 3A is different from 1A 2A (i.e. the cards are distinct)
You can have the empty hand or 0 cards. (if this is wrong and the others are right it would be 80 possibilities)
But if those assumptions are right, this is equivalent to a situation where I have 4 ordered values that each have 3 possibilities, which is 3*3*3*3, or 81.

Maarten D. de Jong
Netherlands Zaandam

ex1st wrote: Would it be better if each card had 3 sides: A, B and N (N meaning Not there)? Better in what way? I don't understand what you're asking here.
Quote: How many combinations are possible? 81? 4C1 * 2^1 + 4C2 * 2^2 + 4C3 * 2^3 + 4C4 * 2^4 = 8 + 24 + 32 + 16 = 80 with ..C.. being the combinatorial function: xCy = choose y things out of a total of x, ignoring the same combination (12 = 21, 123 = 132 = 321 = ..., etc.).
The extra factors come in from the fact that a card has a front and back; and here combinations are not useful. 1B2A is not the same as 1A2B, for example. So for every card in the resulting combination the combinatorial outcome has to be multiplied by the factors listed.

Charles Ward
Japan Matsumoto Nagano

Thanks Sam. And thanks Maarten.
Berinor wrote: It depends on some assumptions, but at first blush I'd say 81 is right and your reframing to A/B/N is the right way to see it.
A few assumptions that didn't come up in your framing:
1A 2A is the same as 2A 1A (i.e. ordering doesn't matter) YES. As far as I have understood, with permutations the position (or order) is important. With combinations, it is not.
Quote: 1A 3A is different from 1A 2A (i.e. the cards are distinct) YES. The cards are different. 1A is not equal to 2A.
Quote: You can have the empty hand or 0 cards. (if this is wrong and the others are right it would be 80 possibilities) YES. 81  1 = 80. Nice!
Quote: But if those assumptions are right, this is equivalent to a situation where I have 4 ordered values that each have 3 possibilities, which is 3*3*3*3, or 81.
So, for my follow up question:
A certain number of the 4 cards, lets say 2 of the cards (numbers 1 and 2), must always be present, but can still be placed showing side A or B.
How do I combine 4 cards, some of which have 2 options, but others have 3 options, to see how many combinations are possible?
Cheers.

Maarten D. de Jong
Netherlands Zaandam

ex1st wrote: A certain number of the 4 cards, lets say 2 of the cards (numbers 1 and 2), must always be present, but can still be placed showing side A or B. Well, that's easy. If two cards always must be present, then you can divide the original 4 cards into two groups. One group is what must be present, and that gives rise to a multiplication of 4 of the total. So the calculation is now 4 * (the number of combinations possible with 0, 1 or 2 remaining cards).
If a card has 3 genuine options (A, B, and C) as opposed to just 2 with the third being 'not there'), then the calculation becomes more cumbersome as the totalling of all possibilities begins to grow unwieldy. There is no remedy for this, I'm afraid... unless you go to the other extreme and assume an unending supply of cards so that the probability of drawing a particular one becomes fixed instead of dependent on others having been chosen or not. But that approach won't work here as the number of cards is simply too small.
Quote: How do I combine 4 cards, some of which have 2 options, but others have 3 options, to see how many combinations are possible? The solution depends on how many 2 and 3way cards you let participate in the problem; there is no general solution. Because there are 8P4 = 1680 permutations possible to create a set of 4 cards out of a total of one twosided 1, one threesided 1, one twosided 2, one threesided 2, ... I'm not going to list them all. I'm sure you understand . (That said, the way in which a particular combination of 2 and 3way cards behaves is of course independent of its printed matter.)

Charles Ward
Japan Matsumoto Nagano

What if we looked at it from a different point of view?
I have different 8 cards. Each card has 2 sides. Only 1 of those cards is optional.
The following equation expresses the possible combinations of 8 cards with 2 options each:
2^8 (2 to the power of 8) = 256
and here is the equation for there being only 7 cards:
2^7 = 128
Does 256 + 128 = the total number of combinations?

Maarten D. de Jong
Netherlands Zaandam

Yes, it does, but that's not the same as your earlier problem, unless of course you want the third way of a 3way card to mean 'not there'.

Charles Ward
Japan Matsumoto Nagano

Last question for the moment
I have different 14 cards. Each card has 2 sides. Only 6 of those cards are optional.
How do I work out the total number of combinations?

Maarten D. de Jong
Netherlands Zaandam

14  6 = 8. So you start out with a 'basic multiplier' of 2^8 = 256. Then calculate the number of possible combinations achieved with 6 cards which is 3^6 (with option 3 being 'card is not there yet is there'). So the total is 256 * 3^6 which Google informs me is 186624.


It might be helpful to think about this as follows:
Consider the number of positions each card could be in.
A required card could be in 2 positions (side A or side B). An optional card could be in 3 positions (side A, side B, or not present). A onesided required card could be in 1 position (side A). A onesided but optional card could be in 2 positions (side A or not present).
Then, multiply together the number of options for each card.
If card #1 is required (2 options), card #2 is optional (3 options), and card #3 is optional but only onesided (2 options), then you have 2 * 3 * 2 = 12 combinations.
If you've got M cards that can be in 2 positions and N cards that can be in 3 positions, then you have 2^M * 3^N combinations.

Charles Ward
Japan Matsumoto Nagano

I AM NOT WORTHY!! Thanks.

Russell Martin
United Kingdom Liverpool Merseyside
Sanctify!

Essentially all of combinatorics (or the counting part of it) comes down to the "multiplication principle" (or "rule of product", if you like) and the "addition principle" (or "rule of sum").
The rest is fancy ways of applying these principles (binomial coefficients, generating functions, inclusion/exclusion, etc.). And you're seeing applications of these rules in the examples that are being discussed here.


