I am designing a card game inspired by Tetra Master, the card game in Final Fantasy IX. It is a game with a different theme altogether, about causing Stars to go Nova by placing Planets and other Bodies in a way that will cause their Mass Points to exceed or reach critical. The difficulties in design for this game is twofold, one, to determine the Mass Points of every Card, two, the harder one, the lines of influence, what we call the Gravity Lines, showing the lines of interaction with other Cards, on the Star or Planet Card, in a total varying combination of 8 directions. So here is where my question comes in, I am thinking of 20 Asteroid Cards in the game, the special ability of Asteroids is that they can stack on top of each other, with the potential of reaching high combined Mass Points. Here is where this question of permutation and combination comes in, each Asteroid has 3 Gravity Lines, and each set of Gravity Lines have to point in a unique direction combined. Is there a way to compute this?
#Permutation-Combination question for a Game Design
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no
it shows now
no worries
I am saying can you see my post which states not mentioned
Yeah
I will discuss about that
ok
If no one comes ping this role in #off-topic and if someone asks tell them A Log said to do it because no one came
Have a great day π
Uhmmmmm...tbh i have been here for 1 year and still idk.Its like multi topic channel
You can get help or discuss topics in that channel
so the number of different combos of arrows is 8 choose 3 or 56, so if you wanted you could have one of each combonation
If you reduced directions to 6 then 20 card work
yes
hmm
ok, can you tell me the different unique combinations?
ok, take it we can orient the cards in any way
or direction
if there are 8 directions and 3 lines, then to represent each you need 56
but what if we can orient the cards in four ways?
and we don't repeat the same combination?
now it is a permutation question right?
i guess
so what's the answer?
24 if order matters
4 factorial
oh i may have misunderstood the question
If there are 8 possible directions and 4 arrows on a card there are 70 possibilities
8 choose 4 is 70, but 4 choose 3 is 4
I don't understand, why 70? what is the calculation?
8!/((8-4)!-4!)
choose 3 not 4 though
why @untold peak?