How many different ways can a group of 6 people be divided into 3 team

I like to think of it in this way:

Make everyone stand in a straight line. The number of ways of doing this is 6!.

Now, first two people from the left make team A, next two make team B and last two make team C.
We have divided 6 people in three teams (A, B and C) and allocated spots to each member in his team (first person, second person). But the three teams are not distinctly named teams so you un-arrange by dividing by 3!.
Also each team doesn't have an arrangement of "first person and second person" so you un-arrange them by dividing by 2! for each team.

6!/(3!*2!*2!*2!) = 15

Answer (C)
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Hi,

1) 2 people can be choosen in 6C2 ways from 6 people = \(6C2=\frac{6!}{4!2!}= 15\)..
2) further 2 can be selected in 4C2 ways from remaining 4 people = \(4C2 = \frac{4!}{2!2!}= 6\)..
3) finally remaining 2 can be selected in 1 way..

Total ways = \(15*6*1 = 90\)ways..

But in these groups repetitions are there....
example - let them be a,b,c,d,e,f...
let a,b be selected in step(1), c,d in step(2) and e,f in step (3)..
Now in another way.. a,b is selected in step(2), c,d in step (3) and e,f in step (3)

Both the above are same but treated as different ways..
so we require to divide our total by ways the groups can be placed = 3!

answer = \(\frac{90}{3!} = \frac{90}{6} = 15\)
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I dont fully understand about the dividing by 3! part.

what about this question?

ten-highschool-boys-gather-at-the-gym-for-a-game-of-basketball-two-te-219483.html

why not divide by 2!?

Can you explain further?

In the Q mentioned by you, we are dividing in two teams so first a team could be selected in 10C5 or in 5C1..
so two place swhere it can be selected and that is why we divide by 2..

But here we are dividing by 3! because any particular team can be selected in 6C2 or 4C2 or 2C1...

say the members are abcdef...
We select in 6C2 = ab...........4C2 = cd..........2C1= ef....
in another case in 6C2 = ab...........4C2 = ef..........2C1= cd....
in another case in 6C2 = cd...........4C2 = ef..........2C1= ab....
in another case in 6C2 = cd...........4C2 = ab..........2C1= ef....
in another case in 6C2 = ef...........4C2 = ab..........2C1= cd....
in another case in 6C2 = ef...........4C2 = cd..........2C1= ab....

so the distribution is the same BUT it has been calculated as 6 different ways....
so we divide total by 6 or 3! ways..
so we divide by (# of groups)!
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Hi,

I understand that 6 people need to be divided into teams of 2 each. And as it is a team, the arrangement inside should make no difference at all.

Hence 6C2 - 6*5/2 = 15 teams. I can't understand as to why you are again considering "4C2 ways from the remaining people". I guess am missing something monumental here, but i just can't get it. Is the logic that out of 6 people after forming teams of 2 people, you are still left with 4 people and hence can form teams of 2 again. This is my guess but i can't understand how can you consider that, as once you have divided them into 15 teams, you are left with nothing to re-divide?

Hi,

if the Q said that we have to choose two people out of 6, then the answer would be straight 6C2..
But here we are talking of 3 teams of 2 each that is why we select 2 out of 6, then another 2 out of remaining 4 and then take remaining 2 as a team..
so these three 6C2, 4C2 and reamining 2, form a set of 3 teams of 2 players each..

Here since order does not matter, we divide it by 3!, so the answer finally comes same as 6C2....
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From your question I am guessing that you are confusing combinations with permutations.
6C2 is combination which is "only selection". Out of 6 people you select 2 in 15 ways. You do not arrange anyone.

You have 6 people: A, B, C, D, E, F

You can select two in 15 ways:
1. A, B
2. A, C
3. A, D
...
...
15. E, F

You do nothing with the leftover 4 people. So in the next step, you select 2 more out of the 4 remaining to make another team. Finally, you are left with 2 people which make the third team.
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1. The number of ways in which mn different items can be divided equally into m groups, each containing n objects and the order of the groups is not important is (mn)!/ (n!^ m )* m!

2. The number of ways in which mn different items can be divided equally into m groups, each containing n objects and the order of the groups is important is (mn)!/ (n!^ m )

Here order is not important so we have to divide and we apply formula 1
Answer is 15 (C)

Please update 3! value to 6 and answer to 15
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The smoothest way to solve this is by thinking it like this:

- We have 6 people (A, B, C, D, E, F) and we want 3 teams. Therefore we will have 3 2-member teams. The final teams will be for example: AB, CD, EF.
- Imagine a 6x6 table with A, B, C, D, E, F horizontally and vertically. Those are the possible 2-member teams.
- You need the half triangle without the mid-boxes. Meaning you don't want AA, BB, CC, DD, EE, FF and you don't want to double count AB and BA etc..
- Therefore 6*6=36. Minus the mid-boxes we have 30. And the bottom half we want is 15.

Hey found the way really nice, but when I tried the same way for below mentioned question, I could not get correct answer of 3
Q- In how many ways can 4 people be divided into 2 groups of 2 people each?
Did I miss something ?


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I don't understand why 15 ways. What I understand is that, when calculating the teams of two people (that if there are 15) we only calculate the number of ways to choose teams of two members, but it is necessary to calculate the number of ways in which you choose three of that number of equipment. I don't know if my idea is correct.

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Thank you for such an interesting approach. This approach works great for questions where certain things have to be distributed equally.

Doubt 1: Will this approach also work for other distributions?

Ex. Distribute 10 different balls in 8 boxes such that boxes have 4,3,2 and 1 balls.

Doubt 2: What do we do in case the things to be distributed are not different but identical. Ex: identical balls

Looking forward to your explanation!

No. Of ways selecting 2 from 6 is 6c2= 15.
Mentioning 3 teams is just to confuse.

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