Solution and formula would be awesome for the following:
1. How many possible combinations of teams could there be if there are 4 players and 2 players per team? Obviously, this would be three:
combination one: 12 34
combination two: 13 24
combinations three: 14 23
2. Same as above except for 6 players and 2 players per team.
3. Same as above except for 8 players and 2 players per team.
4. Same as above except 9 players and 3 players per team.
Thanks for the help.
anyone board?
Use nCr = n!/ r!(n-r)!
n is the number per team.
r is the amount of players.
1. nCr = 4!/ 2!(4-2)! = 24/4 = 6
2. nCr = 6!/ 2!(6-2)! = 30/2 = 15
3. nCr = 8!/ 2!(8-2)! = 56/2 = 28
4. nCr = 9!/ 3!(9-3)! = 504/6 = 84
Yellowbook wrote:
Use nCr = n!/ r!(n-r)!
n is the number per team.
r is the amount of players.
1. nCr = 4!/ 2!(4-2)! = 24/4 = 6
2. nCr = 6!/ 2!(6-2)! = 30/2 = 15
3. nCr = 8!/ 2!(8-2)! = 56/2 = 28
4. nCr = 9!/ 3!(9-3)! = 504/6 = 84
thanks. but the answer to the first one is 3. its not a simple combination computation, although i am sure that formula is a factor.
I dunno what you mean, my mind might be shot right now but if you have 4 different guys, you can make 6 different teams:
alex, buzz, ned, clem
alex-buzz
alex-ned
alex-clem
buzz-ned
buzz-clem
clem-ned
the difference is that the second team of one pair could be the first team of another pair
12 vs 34
13 vs 24
14 vs 23
23 vs 14- same as #3
24 vs 13-same as #2
34 vs 12- same as #1
Yellowbook wrote:
I dunno what you mean, my mind might be shot right now but if you have 4 different guys, you can make 6 different teams:
alex, buzz, ned, clem
alex-buzz
alex-ned
alex-clem
buzz-ned
buzz-clem
clem-ned
yeah exactly, except it would look like this:
alex-buzz vs. ned-clem
alex-ned vs. clem-buzz
alex-clem vs. ned-buzz
so it would be 3 different combos.
what if there were 6 players of 2 player team. for example consider players a,b,d,e,f,g
a-b vs. c-d vs. e-f
a-b vs. c-e vs. d-f
etc.
let x be the number of total players and y be the players per team
the formula is summation (z from 2y to x) (z-1)*((z-y)choose y))
for example choosing 2 from 6 people is 5*(4 choose 2)+4*(3 choose 2)+3*(2 choose 2) which comes out to 45
this works for 2 players per team
for 3 players, i think you replace z-1 with the amount of teams while keeping 1 player in, aka
123,124...189=28
234,235...289=21
345,346...389=15
=10
=6
=3
=1
so the first term of choosing 3 from 9 would be 28(7 choose 3)
the summation might be different, but just stop once z-y=y
pretty sure this is right but its difficult to write math on a computer
I think the following will arrive at the same result:
6 people w/2 per team: (6 choose 2)*(4 choose 2)/2
8 people w/2 per team: (8 choose 2)*(6 choose 2)/2
9 people w/3 per team: (9 choose 3)*(6 choose 3)/2
N people w/n per team: (N choose n)*(N-n choose n)/2
You asses. Only Jesus Christ Our Lord and Savior can determine those sorts of outcomes. No devil Muslim or Zionist or Heathen has a say in such matters.
There are clearly 15 possibilities with 6 people. Here they are:
12 34 56 12 35 46 12 36 45
13 24 56 13 25 46 13 26 45
14 23 56 14 25 36 14 26 35
15 23 46 15 24 36 15 26 34
16 23 45 16 24 35 16 25 34
I think the generalization is this. You can work the problem recursively. For 4 people, there are 3 possiblities. For 6 people, there are five people that person A can be with. Once you know who person A is with, you know from your original reasoning that there are three other possibilities for the remaining 4 people. So five times three equals 15.
To extend this out, for eight people, there are seven possiblities for person A to be with. For the remaining 6 people, there are 15 possibilities as shown above. So 15*7=105, which is your answer.
For ten people, the answer is 105*9=945.
Does this look kosher?????