A group of 6 friends want to play a trivia game. How many

How about the same question, but this time two teams of three people? 6C3??

1st team can be assigned in : 6C2 ways = 15
2nd Team can be assigned in: 4C2 ways = 6
3rd team can be assigned in: 2C2 ways = 1

however, order is not important in team sequences ((1,2),(3,4),(5,6) is same as (3,4),(1,2),(5,6)):

therefore total number of ways = 90 / 3! = 15

quoted



These problems are a slight twist on combination problems. The combination formula can be used to determine the number of ways each group can be formed. The multiplication principle is then used to determine the number of ways the groups can be selected together.

A group of 6 friends want to play a trivia game. How many different ways can the ways can the friends be divided into 3 teams of 2 people?

The order of selection does not matter. Thus, this is a combination problem.

The number of combinations of n objects taken r at a time i

There are C(6,2) ways to select the first team.

C(6,2) = 6!/(2!(4!))
C(6,2) = 720/48
C(6,2) = 15

Since 2 friends have now been selected, there are C(4,2) ways to select the second team.

C(4,2) = 4!/(2!(2!))
C(4,2) = 24/4
C(4,2) = 6

Since 4 friends have now been selected, there are C(2,2) ways to select the third team.

C(2,2) = 1

The Multiplication Principle tells us that the number of ways independent events can occur together can be determined by multiplying together the number of possible outcomes for each event.

15 * 6 * 1 = 90

The order of team selection does not matter. Thus, this is an unordered partition. To account for the unordered partition, divide 90 by the factorial of the number of teams.

= 90/3!
= 90/6
= 15

In your logic after A has chosen 1 of 5 ways.............. B only has 3 ways to choose considering he was not chosen by A.
Your logic seems simple but I am really not 100% sure if its correct.