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There is 3 ways you can choose 2 people from a group of 4. After this choice is made you have two people left (therefore the other team is already chosen)
so there is 3 ways in which you can choose 2vs2 teams out of 4 people.
There is 3 ways you can choose 2 people from a group of 4. After this choice is made you have two people left (therefore the other team is already chosen)
so there is 3 ways in which you can choose 2vs2 teams out of 4 people.
The long way:
Given A,B,C,D
AB Vs CD AC Vs BD AD Vs BC
there is no other way you can choose the teams
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How would you do a more difficult problem such as: In a group of 8 people, how many different 4 versus 4 teams could you form? thx
Can someone tell my why 4C2 * 2C2 = 6, is not the correct answer?
Can someone tell my why 4C2 * 2C2 = 6, is not the correct answer?
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ok lets see 4C2 = the number of ways you can choose 2 from 4 people = 6 (the total number of teams) BUT a team containing a certain teammember cannot play another team containing the same team member, eg AB cannot play AC,AD,BC, BD since A and B are unique.
So out of 6 teams, there can be 6C2 = 15 combinations of 2 teams (to play against each other). BUT since some teams cannot play each other (unless we clone individuals ) we have to divide this number by 5 (the number of teams with either player X or Y)
young_gun
How would you do a more difficult problem such as: In a group of 8 people, how many different 4 versus 4 teams could you form? thx
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in the same was as above 8C4 gives 70 teams. Divide this by 2 gives the 35 possible matchups (i am not 100% sure about this scenario, maybe a math maestro like Walker can shed some light)
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