A group of 8 friends want to play doubles tennis. How many

The first team can be selected in 8c2 ways, the second in 6c2, the third in 4c2 and the fourth in 2c2 ways.

Multiplying the no. of ways, we have
(8!)/16 = 2520 ways.

The answer is B.

There's another topic asking this very question down below. Here's my answer:

E) 105.

Start with the question, how many ways can you divide a group of 4 into 2 teams. The answer is 3. Suppose our group is A B C D. Then the possible teams are:

AB CD
AC BD
AD BC

(It's 4C2/2, because determining one of the teams automatically determines the other one.)

Now consider the case of a group of 6 divided into 3 teams. Our group: A B C D E F. A has to be on some team, and there are 5 possibilities: AB, AC, AD, AE, or AF. For each of these possibilities, the rest of the members form a group of 4 divided into 2 teams. So the overall result is

5*(4C2/2) = 15.

Now we're up to our case. Group = A B C D E F G H. Again, A must be on a team--there are 7 possibilities. For each of those possibilities, there are 15 ways of dividing the remaining 6 members into 3 teams. So total possibilities = 7 * 15 = 105.

8c2 x 6c2 x 4c2 = 2520 = B

Bunuel... I guess I have started to feel lost when it comes to Combinations and Permutations. I thought since I have used Combination, I don't consider the order.. but as we per you even 8c2 x 6c2 x 4c2 has considered the order in it... THis confuses me... Could you please help me with this?

For me I thought.. if I had used 8p2 x 6p2 x 4p2.... I would have considered the order too.... is that not the case?

Thanks a lot Bunuel!

You chose 4! because of 4 sets? right. If we had 5 teams we would have chosen 5! irrespective of the fact that from how many people its being chosen from?

Yes we are dividing by the factorial of the number of teams. You can also check following problems to practice:

probability-88685.html#p669025
sub-committee-86346.html#p647698
probability-85993.html#p644656
combination-groups-and-that-stuff-85707.html#p642634

Hope it helps.

Thanks for these superb posts! The next difficult part is how to judge whether the order is important or not? Can you shed some light?

In the first post on your list of suggested posts, why is order important for 9 dogs to be divided into 3 groups of 3 members each?

The order of the groups matter in case we have team #1, #2, etc. Or in other words when we have specific assignment for each team.

Consider the following: we have one set of three teams: A, B & C and three tasks: 1, 2, & 3. In how many ways can we assign these teams to these tasks? The answer would be 3!=6:

1-2-3
A-B-C
A-C-B
B-A-C
B-C-A
C-A-B
C-B-A

Ian Stewart explains this very well at: combination-groups-and-that-stuff-85707.html#p642634

Hope it helps.

Hi how do we get the answer for the Ist question as 105. Pls explain

A group of 8 friends want to play doubles tennis. How many different ways can the group be divided into 4 teams of 2 people?

A. 420
B. 2520
C. 168
D. 90
E. 105

\(\frac{C^2_8*C^2_6*C^2_4*C^2_2}{4!}=105\), we are dividing by 4! (factorial of the # of teams) as the order of the teams does not matter. If 8 people are - 1, 2, 3, 4, 5, 6, 7, 8, then (1,2)(3,4)(5,6)(7,8) would be the same 4 teams as (5,6)(7,8)(1,2)(3,4), as we don't have team #1, team #2...

You can think about this in another way.
For the first person we can pick a pair in 7 ways;
For the second one in 5 ways (as two are already chosen);
For the third one in 3 ways (as 4 people are already chosen);
For the fourth one there is only one left.

So we have 7*5*3*1=105

You can check similar problems:
https://gmatclub.com/forum/probability-8 ... %20equally
https://gmatclub.com/forum/probability-8 ... ide+groups
https://gmatclub.com/forum/combination-5 ... ml#p690842
https://gmatclub.com/forum/sub-committee ... ide+groups


There is also direct formula for this:

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 \(\frac{(mn)!}{(n!)^m*m!}\).

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 \(\frac{(mn)!}{(n!)^m}\)

Hope it helps.

Formula = 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 important is (mn)!/n^m*m!

so its 8!/(2^4*4!) = 105 = E

From the question, we can say that the teams are not distinct i.e. we don't have team A, team B etc. But let's solve this question by making first team, second team, third team and fourth team. Later we will adjust the answer.
Out of 8 people, how can you make the first team? In 8C2 ways.
Out of 6 people, how can you make the second team? In 6C2 ways.
Out of 4 people, how can you make the third team? In 4C2 ways.
Out of 2 people, how can you make the fourth team? In 2C2 ways.
How can you make the 4 teams?
8C2 *6C2*4C2*2C2

But here, we have considered the 4 teams to be distinct. We called them 'first team', 'second team' etc.
So we need to divide the result by 4! to 'un-arrange' them.

You get: 8C2 *6C2*4C2*2C2/4! = 8*7*6*5*4*3*2*1/16*4! = 105

karishma,




Can you please explain this?
I understand that we divide the slots! to remove identical stuff but here how does it make sense?

This is the logic behind this step:

Say there are 4 boys: A, B, C, D
There are two ways of splitting them in two groups.

Method I

The two groups can be made in the following ways
1. AB and CD
2. AC and BD
3. AD and BC

The groups are not named/distinct. You have 4 boys in front of you and you split them in 2 groups and do not name the groups. There are 3 total ways of doing this.

Method II
On the other hand, I could put them in two distinct groups in the following ways
1. Group1: AB, Group2: CD
2. Group1: CD, Group2: AB (If you notice, this is the same as above, just that now AB is group 2)
3. Group1: AC, Group2: BD
4. Group1: BD, Group2: AC
5. Group1: AD, Group2: BC
6. Group1: BC, Group2: AD

Here I have to put them in two different groups, group 1 and group 2. AB and CD is not just one way of splitting them. AB could be assigned to group 1 or group 2 so there are 2 cases. In this case, every 'way' we get above will have two possibilities so total number of ways will be twice.
So there will be 6 total ways.

Here since the groups are not distinct but 8C2 * 6C2 * 4C2 * 2C2 makes them distinct (we say, select the FIRST group in 8C2 ways, SECOND group in 6C2 ways etc), we need to divide by 4!.

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 \(\frac{(mn)!}{(n!)^m*m!}\).

How many different ways can the group be divided into 4 teams (m) of 2 people (n)?

\(\frac{(mn)!}{(n!)^m*m!}=\frac{8!}{(2!)^4*4!}=105\).

Hope it helps.

I do not agree that we need to divide it by 4!

8C2 * 6C2*4C2 is a regular combination formula

We choose 2 persons out of 8. 1st team is formed. There is no arrangement, just selection. (8C2)

We then choose 2 persons out of remaining 6. 2nd team is formed. There is no arrangement, just selection. (6C2)

We then choose 2 persons out of remaining 4. 3rd team is formed. There is no arrangement, just selection. (4C2)

We then choose 2 persons out of remaining 2. 4th team is formed. There is no arrangement, just selection. (2C2)

Product: 8C2 * 6C2*4C2

This works out to 8! / (2!.2!.2!.2!) = 7!/2 = 2520