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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

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

We should divide this by 4! --> 2520/4!= 105, as the order of the teams does not matter.

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.

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

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

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.

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?

It's not about the order of the players IN the group, with C you you excluded the duplications of these kind: {AB} and {BA} are excluded by C.

It's about the order of the groups themselves. Set of teams: {AB}{CD}{EF}{GH} is the same set as: {E,F}, {C,D}, {A,B}, {G,H} and by your formula we'll get these two identical sets of teams.

Basically one particular set of teams, let's say again: {AB}{CD}{EF}{GH}, will be chosen by your formula 24 times (4!=24) and we "need" only one from these 24, hence we should divide 8c2 x 6c2 x 4c2 by 4! to get rid of these duplications.

It's not about the order of the players IN the group, with C you you excluded the duplications of these kind: {AB} and {BA} are excluded by C.

It's about the order of the groups themselves. Set of teams: {AB}{CD}{EF}{GH} is the same set as: {E,F}, {C,D}, {A,B}, {G,H} and by your formula we'll get these two identical sets of teams.

Basically one particular set of teams, let's say again: {AB}{CD}{EF}{GH}, will be chosen by your formula 24 times (4!=24) and we "need" only one from these 24, hence we should divide 8c2 x 6c2 x 4c2 by 4! to get rid of these duplications.

Hope it's clear.

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:

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

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.

Nice explanation....Could you please explain to me where i got it wrong?...

Walker, can you please explain why you are dividing 8C2 * 6C2 * 4C2 by 4!
i seem to understand 7*5*3 approach, but want to know your reasoning. thanks!

Walker, can you please explain why you are dividing 8C2 * 6C2 * 4C2 by 4! i seem to understand 7*5*3 approach, but want to know your reasoning. thanks!

4!=4P4 - we should exclude arrangement of 4 pairs in different order.

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

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

Cheers! JT........... If u like my post..... payback in Kudos!!

|Do not post questions with OA|Please underline your SC questions while posting|Try posting the explanation along with your answer choice| |For CR refer Powerscore CR Bible|For SC refer Manhattan SC Guide|

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

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

We should divide this by 4! --> 2520/4!= 105, as the order of the teams does not matter.

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.

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

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? _________________

Cheers! JT........... If u like my post..... payback in Kudos!!

|Do not post questions with OA|Please underline your SC questions while posting|Try posting the explanation along with your answer choice| |For CR refer Powerscore CR Bible|For SC refer Manhattan SC Guide|

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?

It's not about the order of the players IN the group, with C you you excluded the duplications of these kind: {AB} and {BA} are excluded by C.

It's about the order of the groups themselves. Set of teams: {AB}{CD}{EF}{GH} is the same set as: {E,F}, {C,D}, {A,B}, {G,H} and by your formula we'll get these two identical sets of teams.

Basically one particular set of teams, let's say again: {AB}{CD}{EF}{GH}, will be chosen by your formula 24 times (4!=24) and we "need" only one from these 24, hence we should divide 8c2 x 6c2 x 4c2 by 4! to get rid of these duplications.

Hope it's clear.

Thanks a lot Bunuel! _________________

Cheers! JT........... If u like my post..... payback in Kudos!!

|Do not post questions with OA|Please underline your SC questions while posting|Try posting the explanation along with your answer choice| |For CR refer Powerscore CR Bible|For SC refer Manhattan SC Guide|

It's not about the order of the players IN the group, with C you you excluded the duplications of these kind: {AB} and {BA} are excluded by C.

It's about the order of the groups themselves. Set of teams: {AB}{CD}{EF}{GH} is the same set as: {E,F}, {C,D}, {A,B}, {G,H} and by your formula we'll get these two identical sets of teams.

Basically one particular set of teams, let's say again: {AB}{CD}{EF}{GH}, will be chosen by your formula 24 times (4!=24) and we "need" only one from these 24, hence we should divide 8c2 x 6c2 x 4c2 by 4! to get rid of these duplications.

Hope it's clear.

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?

It's not about the order of the players IN the group, with C you you excluded the duplications of these kind: {AB} and {BA} are excluded by C.

It's about the order of the groups themselves. Set of teams: {AB}{CD}{EF}{GH} is the same set as: {E,F}, {C,D}, {A,B}, {G,H} and by your formula we'll get these two identical sets of teams.

Basically one particular set of teams, let's say again: {AB}{CD}{EF}{GH}, will be chosen by your formula 24 times (4!=24) and we "need" only one from these 24, hence we should divide 8c2 x 6c2 x 4c2 by 4! to get rid of these duplications.

Hope it's clear.

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:

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:

Thanks for sharing your simple n elegant approach.

Bunuel wrote:

jeeteshsingh wrote:

Balvinder wrote:

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

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

We should divide this by 4! --> 2520/4!= 105, as the order of the teams does not matter.

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.

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

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