I have a rather simple math question:
If I have five (5) colors and use them four (4) times, how do I compute the UNIQUE number of combinations.
An example:
If I choose two (2) colors = Green and Black. I get duplicates like:
GGKK = GKGK = KGKG = GKKG = KGGK or
GKKK = KGKK = KKGK = KKKG
Can somebody help me in knowing how to mathematically calculate the UNIQUE count???
If you're looking to get rid of any replicates, a combination would be the way to go.
http://www.mathsisfun.com/combinatorics/combinations-permutations.html
Do you have duplicates of your colors when choosing, as your example above? If you do, it might require a bit more work, but I can look into it further.
When you say 5 colors, do you mean 5 different colors (aka grabbing from RYGBW out of a bag, for instance, with one of each color in the bag, or are there multiples of each color? if so, how many multiples are there of each color, potentially?
It took me a bit of tinkering but I think this is right: if you have N colors and you want to take K of them, there are "N+K-1 choose K" = (N+K-1)! / (K! * (N-1)!) possibilities. Here ! means factorial, so K! = K * (K-1) * ... * 2 * 1.
So in your case the answer would be 8! / (4! * 4!) = 70 possibilities.
I was looking at that one as well. I think that assumes that you have enough of each color to draw them an infinite amount of times, but it should work if there are enough of each color being chosen from.
There are five (5) colors and I want to compute UNIQUE combinations of four (4) of them. So my colors are: Red (R), Green (G), Blue (B), White (W) and Black (K).
So there are MULTIPLEs of each color. so four (4): Green, Green, Green, Green is okay. And Black, Black, Black, Black is also okay.
But when you do Green + Black (mixed), you need to remove combinations so that they are unique. So Green, Green, Black, Black is okay.
BUT KKGG, GKGK, KGKG, KGGK, GKKG are NOT okay.
Because it works out to be 2x two (2) colors.
Whoa thanks - that is quite the formula. Would have never figured that one out myself...
Much gratitude SLiV and JTAshby!!!
I knew that 5 x 5 x 5 x 5 = 625 is the total amount of combinations.
What I could not figure out, is how to remove the duplicates...
Thanks again!
BTW does anyone know a site that can allow you to compute all 70 possibilities. I know 70 isn't HUGE - but it's quite a bit - I may try to compute them manually and see how that turns out.
Keep you all posted!
http://www.mathsisfun.com/combinatorics/combinations-permutations-calcul...
save you a bit of time, if you haven't done it already...
With my software (nanDECK) non only you can compute the number of combinations, but you can use them in drawing the cards.
In your example, this line creates a sequence of all the combinations (order doesn’t matter) of five elements, taken by groups of four (there are 70 results):
cr[list]4=R|G|B|W|K
This line is for the permutations (order matters), there are 625 results:
pr[list]4=R|G|B|W|K
The same, but removing the same results (these are 165):
prx[list]4=R|G|B|W|K
And with these two lines, you can print them on a deck of cards:
font=arial,32,,#000000
text=1-{(list)},[list],0,0,100%,100%
Download page for nanDECK:
http://www.nand.it/nandeck
 | Board Game Blueprint - New Episode Every Wednesday (21) by The Game Crafter |
| Premium Bullet & Premium Toxic Waste Board Game Pieces at The Game Crafter (0) by The Game Crafter |
| Build your own [insert game genre here] (3) by larienna |
 | PoA — Major shift back closer to FCE (2) by questccg |
| What “Should” Be in an RPG Design Book (11) by lewpuls |
| Blank Poker Card Sale - 3 Cents Each! (0) by The Game Crafter |
| Blank Playing Cards - Bridge 57mm x 89mm UK (1) by questccg |
| Finally returned after all these years (1) by DyminoMonsters2004 |
| State of the let-off Union - November 2024 (0) by let-off studios |
| Shoppe: The Simulation of Guilds (1) by questccg |
| The fine line between a game and a simulation (22) by X3M |
| Only 24 hours left to bid on games for the Extra Life Charity Auction (0) by The Game Crafter |
| Songs of Conquest is now 60% off plus an additional discount for... (5) by questccg |
| Returned the reMarkable 2 and purchased the BOOX Go 10.3 (3) by questccg |
| Happy Halloween 2024 (0) by questccg |
| Epic Metal Monster Coins - Now on Kickstarter - Created by The Game Crafter (0) by The Game Crafter |
| DuelBotz: Sample New Card (12) by questccg |
| 2 levels for an unit (wargames) (6) by X3M |
| Dragon Spark Playthrough (0) by The Game Crafter |
| New Board Game Pieces - Premium Water Droplet & Premium Blood Droplet (0) by The Game Crafter |
| Designer with an 'almost' ready product (18) by questccg |
| Protospiel Madison - Only 17 Days Away! (0) by The Game Crafter |
| New Board Game Pieces - Premium Milk Bottle & Premium Beer Mug (0) by The Game Crafter |
| Testing chat GPT for mechanics searching (6) by larienna |
| Epic Metal Monster Coins - Coming soon to Kickstarter - Need your feedback! (2) by questccg |
Well duplicates like Green-Green-Black-Black is still identical to Black-Black-Green-Green.
In BOTH cases there are 2 Green and 2 Black.
Well I have five (5) colors (n=5) and want to pick four (4) (r=4) of them.
BUT I don't know HOW to eliminate duplicates (as I mentioned above) :(
This is a *Tricky* problem...