Color- All 3 are different
Shading- All 3 are different
Number- All are the same; they have 2 numbers
Shape- All are the same; oval
Color- All are the same; red
Shading- All are the same; blank
Number- All are different
Shape- All 3 are different
Color- All 3 are different
Shading- All are the same; stripes
Number- All 3 are different
Shape- All 3 are different
Color- All are the same; they are purple
Shading- All 3 are different
Number- All are the same; they have 3 numbers
Shape- All 3 are different
Color- All 3 are different
Shading- All 3 are different
Number- All are the same; they have 3 numbers
2 of the 4 characteristics can be represented, with the 2 being shading and number. The shape and color stays the same.The first idea that comes to mind is finding the maximum amount with 9 cards; that fits nicely with the 3 different characteristics in each group. I think it would look something like this:
So, if we have these 9 cards, there would be 12 sets. Seems like we will have 12 sets when we have 12 cards, right?
Well, 12 cards will be harder to account for. There aren't 4 characteristics in any of the groups, making it difficult to choose which cards to use for the remaining 3 spots.
And, what's worse, we can't really just add on something like this, because it wouldn't work well with the red.
I think there is one better option than this;
I believe that this is the best thing you can add on to the 9 cards, it creates 2 extra sets instead of 1. I'll show how the sets are made.
This should be the best 12 card game.
The 2 new sets that would be created are these:
These sets kind of solve the problem of having only red and green cards. We add on a purple card with 3 shapes instead of 2, because if it was purple with 2 shapes, it would only create 1 set.
Now, we can create cards each with 1, 2, and 3 shapes, and different colors.
The reason why we don't have a set with the 1st card being a striped red diamond and the 2nd card being a solid green card is because we don't have enough cards to add on; we only have 3 extra cards to add, even though we really want to add 4. So, if we had a 13 card game, we would be able to get an extra set added on.
So, I believe that a 12 card game can have a maximum of 11 possible sets, if all of the cards are different.
So, it looks like we're all done!
Well, there are a few more things we could do...
What if a set could have cards be the same as each other? What if every single card was the same??? How many sets could be made?
Here's an example of having every single card be the same:
I know. It looks absolutely crazy hard to understand.
No sarcasm whatsoever at all.
This is actually pretty easy; there are 12 cards to choose from at the beginning, and after selecting your first card, you have 11 cards to choose from, and 10 cards which could represent the last card.
So, really what we're doing is 12*11*10, or just multiplying together all the possibilities of picking the 12, 11, and 10 cards available, resulting with 1320 possible combinations.
There was actually a formula created to do this easier, called the permutation formula.
This is the formula, where N is the number of cards and R is the number of things you are choosing from.
P(n,r) = n!/(n-r)!
So, if N is 12 and R is 3,
just put it into the formula, and solve.
P(12,3) = (12!)/((12-3)!)
P(12,3) = (12!)/(9!)
P(12,3) = (12*11*10*9!)/(9!)
P(12,3) = 12*11*10
P(12,3) = 1320