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

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Manager
Joined: 11 Jul 2010
Posts: 210
A group of 8 friends want to play doubles tennis. How many  [#permalink]

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14 Nov 2010, 04:18
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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

Hypo:

Instead, if the problem were to say: A group of 9 friends want to play doubles tennis. How many different ways can the group be divided into 4 teams of 2 people?

Can the answer be arrived at as follows?

select 8 out of 9 people, first: 9C8 = 9

Then select the sets of 2 people

8!/2^4*4! = 105

answer = 9*105 ---> is this correct? can anyone who knows tell me? Many thanks.
Math Expert
Joined: 02 Sep 2009
Posts: 47157
A group of 8 friends want to play doubles tennis. How many  [#permalink]

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15 Nov 2010, 01:57
3
7
AkritiMehta wrote:
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...

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:
http://gmatclub.com/forum/probability-8 ... %20equally
http://gmatclub.com/forum/probability-8 ... ide+groups
http://gmatclub.com/forum/combination-5 ... ml#p690842
http://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.
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14 Nov 2010, 09:50
gmat1011 wrote:
Disclaimer: This is a hypothetical set up for purposes of understanding the concept. I don't have a definitive answer, so don't sue me

Known Problem: discussed on this forum a few times in the past
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

Hypo:

Instead, if the problem were to say: A group of 9 friends want to play doubles tennis. How many different ways can the group be divided into 4 teams of 2 people?

Can the answer be arrived at as follows?

select 8 out of 9 people, first: 9C8 = 9

Then select the sets of 2 people

8!/2^4*4! = 105

answer = 9*105 ---> is this correct? can anyone who knows tell me? Many thanks.

So the answer to Q1 is 105

And then what you are saying is you choose the one person who sits out (9 ways) TIMES the number of ways to form the teams

This approach is correct
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14 Nov 2010, 22:08
Hi how do we get the answer for the Ist question as 105. Pls explain
Manager
Joined: 11 Jul 2010
Posts: 210

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15 Nov 2010, 06:45
Thanks Shrouded for confirming that.

Thanks Bunuel.
Manager
Joined: 13 Jul 2010
Posts: 149

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16 Nov 2010, 19:12
Bunuel wrote:
AkritiMehta wrote:
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

$$\frac{C^2_8*C^2_6*C^2_2}{4!}=105$$, 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...

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:
probability-88685.html?hilit=different%20items%20divided%20equally
probability-85993.html?highlight=divide+groups
combination-55369.html#p690842
sub-committee-86346.html?highlight=divide+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.

Bunuel - the formula you cite above is this a derivation of the standard combinations formula? I have never seen the exponent in the denominator? Thanks.
Senior Manager
Joined: 18 Jan 2010
Posts: 254
Re: A group of 8 friends want to play doubles tennis. How many  [#permalink]

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27 May 2016, 23:20
Bunuel wrote:
AkritiMehta wrote:
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

$$\frac{C^2_8*C^2_6*C^2_2}{4!}=105$$, 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...

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:
probability-88685.html?hilit=different%20items%20divided%20equally
probability-85993.html?highlight=divide+groups
combination-55369.html#p690842
sub-committee-86346.html?highlight=divide+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.

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
Math Expert
Joined: 02 Aug 2009
Posts: 6258
Re: A group of 8 friends want to play doubles tennis. How many  [#permalink]

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28 May 2016, 00:10
1
1
Bunuel wrote:
AkritiMehta wrote:
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

$$\frac{C^2_8*C^2_6*C^2_2}{4!}=105$$, 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...

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:
probability-88685.html?hilit=different%20items%20divided%20equally
probability-85993.html?highlight=divide+groups
combination-55369.html#p690842
sub-committee-86346.html?highlight=divide+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.

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

Hi,
the solution requires that the answer be divided by 4!...
reason is to eliminate the repetitions that arise out of the standard formula...
Explanation of the same in a simpler scenario..

say 4 people, ABCD in 2 groups...
standard formula $$= 4C2*2C2 = 6*1 = 6$$....
write these cases-
1) AB and CD
2) AC and BD

so only three ! where are other three?
say in 4c2 we chose BC , so AD in second group..
so BC and AD... BUT this is SAME as 3 above - AD and BC..
so when we choose AD, we are automatically choosing the other group BC..
so our answer $$= \frac{6}{2!}= 3$$

Hope it helps you
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Joined: 18 Jan 2010
Posts: 254
A group of 8 friends want to play doubles tennis. How many  [#permalink]

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28 May 2016, 00:35
Thanks for explanation. I would say that there is an obvious ambiguity in the language of the question. If you see that (in your example) the number of groups that you formed are still 6. That is what the question has asked. Infact after selection you are taking arrangements into account and arrangements work out to be similar and you apply correction for the same. I am fine with it.

My limited point is that the question does not adequately clarifies that it is an arrangement question in the way it is phrased.

Because of this ambiguity, I believe that this question is unlikely to be a GMAT OG / GMAT prep / Real GMAT problem.
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Re: A group of 8 friends want to play doubles tennis. How many  [#permalink]

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13 Jun 2017, 12:13
1
1
gmat1011 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

First, we need to select the groups:

Since we have 8 people, the first team can be formed in 8C2 = (8 x 7)/2! = 28 ways. Since there are now 6 people left, the second team can be selected in 6C2 = (6 x 5)/2! = 15 ways. Since there are 4 people left, the third team can be selected in 4C2 = (4 x 3)/2! = 6 ways. Since there are 2 people left, the final team can be selected in 2C2 = 1 way.

Thus, the total number of ways to select the 4 teams is 28 x 15 x 6 x 1 = 2520, if the order of selecting these teams matters. However, the order of selection doesn’t matter, so we have to divide by 4! = 24. Thus, the total number of ways to select the teams when order doesn’t matter is 2520/24 = 105.

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Re: A group of 8 friends want to play doubles tennis. How many  [#permalink]

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21 Aug 2017, 00:18
gmat1011 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

Hypo:

Instead, if the problem were to say: A group of 9 friends want to play doubles tennis. How many different ways can the group be divided into 4 teams of 2 people?

Can the answer be arrived at as follows?

select 8 out of 9 people, first: 9C8 = 9

Then select the sets of 2 people

8!/2^4*4! = 105

answer = 9*105 ---> is this correct? can anyone who knows tell me? Many thanks.

We need to select 4 teams of group of 2 peoples..
Total 8 people are there.

So, No. of ways of selecting 1st group = 8C2 = 28
No. of ways of selecting 2nd group = 6C2 = 15
No. of ways of selecting 3rd group = 4C2 = 6
No. of ways of selecting 4th group = 2C2 = 1

So, if we have to find the number of ways keeping in view that 1st,2nd, 3rd and 4th groups are specific like representing their countries..
But here this is not the case. Groups are equivalent and we don't need to consider the arrangements of group .

So, Final number of ways in which the group can be divided into 4 teams of 2 people = 28*15*6*1/4! = (28 *15*6) /(4*3*2) = 105

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Re: A group of 8 friends want to play doubles tennis. How many  [#permalink]

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22 Aug 2017, 03:47
1
Bunuel wrote:
AkritiMehta wrote:
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...

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:
http://gmatclub.com/forum/probability-8 ... %20equally
http://gmatclub.com/forum/probability-8 ... ide+groups
http://gmatclub.com/forum/combination-5 ... ml#p690842
http://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.

Hi Bunuel
I guess there is a slight typo error . All the combinations should be C28∗C26∗C24∗C224!=105C82∗C62∗C42∗C224!=105
C^2_4* is missing i guess.

kudos if my post helped .
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Posts: 47157
Re: A group of 8 friends want to play doubles tennis. How many  [#permalink]

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22 Aug 2017, 03:49
sandysilva wrote:
Bunuel wrote:
AkritiMehta wrote:
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...

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:
http://gmatclub.com/forum/probability-8 ... %20equally
http://gmatclub.com/forum/probability-8 ... ide+groups
http://gmatclub.com/forum/combination-5 ... ml#p690842
http://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.

Hi Bunuel
I guess there is a slight typo error . All the combinations should be C28∗C26∗C24∗C224!=105C82∗C62∗C42∗C224!=105
C^2_4* is missing i guess.

kudos if my post helped .

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Edited. Thank you.
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Re: A group of 8 friends want to play doubles tennis. How many &nbs [#permalink] 22 Aug 2017, 03:49
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