In how many ways can a group of 6 people be split into three teams of

Total cases: 6!

Now we have to discard when in first place and in second place we have the same people exchanged.
For example: let's name people 1,2,3,4,5,6
one group combination could be:
12, 34, 56
another could be:
21, 34, 56
but this is the same case as the first one. The same occurs with the second and third group. This is why it must be divided by 2!*2!*2!.

Now we have to discard the order among groups.
For example in the 6! cases:
12, 34, 56
is different to
34, 12, 56
when it isn't in our question because the order among groups does not matter. So you have to divide by 3P3 which is the number of permutations of 3 groups grouped from 3 to 3.

Total: \(\frac{6!}{2!*2!*2!*3!} = 15\)

@MathRevolution..why have you multiplied 3!?

I think it because 6! / (2!*2!*2!) is the same as 6C2*4C2*2C2 and because whenever we use combination formula like this for the same group of objects or people we are making an order in the selection process of these 3 groups, though it is only combination formula that we use.

or In other words 6!, which is 6P6, when divide by 2! * 2! * 2!, we are only eliminating the ordering between the two objects within each of the three groups; And in order to eliminate the ordering among the 3 groups themselves, we need to divide again by 3!.

Not sure if you are still looking for an answer for this, but thought this post will help at least others who are confused, as I was till a few hours back.

Method-1:

Since three teams must have equal members therefore each team must have \(6÷3 = 2\) members
First team may be picked out of 6 members in \(^6C_2\) ways
Second team may be picked out of remaining 4 members in \(^4C_2\) ways
Third team may be picked out of remaining 2 members in \(^2C_2\) ways

Total ways = \(^6C_2\)*\(^4C_2\)*\(^2C_2\) But this result included Arrangements of 3 teams with each other in 3! ways

therefore total ways of making three distinct teams = \(\frac{^6C_2 * ^4C_2* ^2C_2 }{ 3!} = 15\) ways

Answer: Option C

Method-2:

Since three teams must have equal members therefore each team must have \(6÷3 = 2\) members
First member of first team may be picked in 8 ways
Second member of first team may be picked in 7 ways
Total ways of making first team = 8*7
But this result includes arrangement between those two chosen members hence we exclude the arrangements by dividing the result by 2! ways
i.e. Ways of making first team \(= \frac{6*5}{2!}\) ways

Similarly, i.e. Ways of making Second team \(= \frac{4*3}{2!}\) ways

Similarly, i.e. Ways of making Third team \(= \frac{2*1}{2!}\) ways

I..e. total teams \(= \frac{6*5}{2!}*\frac{4*3}{2!}*\frac{2*1}{2!}\)

But this result includes the arrangement among teams (which may happen in 3! ways for 3 teams) which may be excluded by dividing the above result by 3! ways

i.e. total DIstinct teams \(= [\frac{6*5}{2!}*\frac{4*3}{2!}*\frac{2*1}{2!}] ÷ 3! = 15\)

Answer: Option C

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There is a formula to remember but once we do, it takes a matter of seconds to solve.

m teams containing n people in each team

Formula to divide (mn) people by m teams = \(\frac{(mn)!}{(n!)^m . m!}\)

Total 6 people => 3 teams; 2 people in each team

So, n=2, m=3

\(\frac{(mn)!}{(n!)^m . m!}\) = \(\frac{(3.2)!}{(2!)^3 . 3!}\) = \(\frac{6!}{(2!)^3 . 3!}\) = \(15\)

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