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How many different ways to play doubles tennis ?

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How many different ways to play doubles tennis ?  [#permalink]

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New post 12 Jul 2011, 02:20
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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
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Re: How many different ways to play doubles tennis ?  [#permalink]

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New post 12 Jul 2011, 02:26
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Alchemist1320 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


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
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Re: How many different ways to play doubles tennis ?  [#permalink]

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New post 12 Jul 2011, 03:42
2
Alchemist1320 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


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
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Re: How many different ways to play doubles tennis ?  [#permalink]

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New post 08 Aug 2011, 07:16
sudhir18n wrote:
Alchemist1320 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


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


super! thanks for formula!
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Re: How many different ways to play doubles tennis ?  [#permalink]

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New post 14 Sep 2011, 14:20
karishma,

Quote:
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.



Can you please explain this?
I understand that we divide the slots! to remove identical stuff but here how does it make sense?
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Re: How many different ways to play doubles tennis ?  [#permalink]

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New post 14 Sep 2011, 22:07
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shankar245 wrote:
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!.
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Re: How many different ways to play doubles tennis ?  [#permalink]

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New post 14 Sep 2011, 22:41
thanks karishma for that detailed explanation :)
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Re: How many different ways to play doubles tennis ?  [#permalink]

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New post 15 Sep 2011, 08:57
Thanks Sudhir , I did not (know) the formula .. and was coming up with 8C2 + 6C2+ 4C2 + 2C2 which is obviously wrong. After I read the formula, it came back to me from school days!!

Thanks Karishma, for the detailed explanation it was really helpful.
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Re: How many different ways to play doubles tennis ?  [#permalink]

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Re: How many different ways to play doubles tennis ?   [#permalink] 07 Apr 2019, 01:49
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