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

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Re: A group of 8 friends want to play doubles tennis. How many [#permalink] New post   28 May 2016, 00:10
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adiagr 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

\(\frac{C^2_8*C^2_6*C^2_2}{4!}=105\), 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:
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
3) AD and BC

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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Re: A group of 8 friends want to play doubles tennis. How many [#permalink] New post   13 Jun 2017, 12:13
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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.

Answer: E
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Re: A group of 8 friends want to play doubles tennis. How many [#permalink] New post   10 Dec 2017, 11:06
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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


Let the 8 people be: A, B, C, D, E, F, G, and H

Take the task of creating the teams and break it into stages.

Stage 1: Select a partner for person A
There are 7 people to choose from, so we can complete stage 1 in 7 ways

ASIDE: There are now 6 people remaining. Each time we pair up two people (as we did in stage 1), we'll next focus on the remaining person who comes first ALPHABETICALLY.
For example, if we paired A with B in stage 1, the remaining people would be C, D, E, F, G and H. So, in the next stage, we'd find a partner for person C.
Likewise, if we paired A with E in stage 1, the remaining people would be B, C, D, F, G and H. So, in the next stage, we'd find a partner for person B.
And so on...

Stage 2: Select a partner for the remaining person who comes first ALPHABETICALLY
There are 5 people remaining, so we can complete this stage in 5 ways.

Stage 3: Select a partner for the remaining person who comes first ALPHABETICALLY
There are 3 people remaining, so we can complete this stage in 3 ways.

Stage 4: Select a partner for the remaining person who comes first ALPHABETICALLY
There is 1 person remaining, so we can complete this stage in 1 way.

By the Fundamental Counting Principle (FCP), we can complete all 4 stages (and thus create 4 pairings) in (7)(5)(3)(1) ways (= 105 ways)

Answer: E

Note: the FCP can be used to solve the MAJORITY of counting questions on the GMAT. So, be sure to learn it.

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Re: A group of 8 friends want to play doubles tennis. How many [#permalink] New post   13 Oct 2018, 17:59
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saurabhgoel 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


The first team can be selected in 8C2 = (8 x 7)/2! = 28 ways.

The next team can be selected in 6C2 = (6 x 5)/2! = 15 ways.

The next team can be selected in 4C2 = (4 x 3)/2! = 6 ways.

The final team can be selected in 2C2 = 1 way.

However, since ORDER OF THE TEAMS DOES NOT MATTER, we need to divide the total number of ways to select the teams by 4! since we have 4 different teams. So we have:

(28 x 15 x 6)/4! = 2520/24 = 105

Answer: E
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Re: A group of 8 friends want to play doubles tennis. How many [#permalink] New post   18 May 2019, 14:26
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


Official Solution


Credit: Veritas Prep

It is quite clear here that the teams are not distinct i.e. we don’t have team 1, team 2 etc. But let’s solve this question by first making team 1, team 2, team 3 and team 4. Later we will adjust the answer.

Out of 8 people, in how many ways can we make team 1? In 8*7/2! ways (i.e. 8C2).
Out of 6 people, in how many ways can we make team 2? In 6*5/2! ways (i.e. 6C2).
Out of 4 people, in how many ways can we make team 3? In 4*3/2! ways (i.e. 4C2).
Out of 2 people, in how many ways can we make team 4? In 2*1/2! ways (i.e. 2C2).

In how many ways can we make the 4 teams? In 8*7*6*5*4*3*2*1/(2!*2!*2!*2!) = 8!/(2!*2!*2!*2!) ways. But here, we have considered the 4 teams to be distinct. Since the teams are not distinct, we will just divide by 4!

We get 8!/(2!*2!*2!*2!*4!) = 105

ANSWER: E
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