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Doubles

by sankruth » Wed Jan 23, 2008 6:59 am
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


Can someone please explain the solution to this problem?
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Re: Doubles

by gabriel » Wed Jan 23, 2008 8:05 am
sankruth 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


Can someone please explain the solution to this problem?
Well.. use combinations, the first team can be selected in 8C2 ways, the second team can be selected in 6C2 ways and the other 2 teams can be selected in 4C2 and 2C2 ways..

So the answer is 8C2*6C2*4C2*2C2 = 8!/(2!)^4 =2520

Regards :)
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answer

by jeet86 » Wed Jan 23, 2008 8:06 am
(b)


8c2*6c2*4c2
=28 * 15 * 6
=2520
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by sankruth » Wed Jan 23, 2008 8:33 am
I got this problem from the set of difficult Math questions in BTG. The answer provided in that document is 105.

Also the reason I feel 2520 is not the correct answer is because the question is asking how many ways the group can be divided into 4 teams of 2 each, not how many different teams can be formed which will be 2520. So the answer mut be less than 2520 but I dont know how to solve it.

For eg. If 8 members are A, B, C, D, E, F, G & H, then A-B, A-C will be a part of 2520 pairs but A-B and A-C cannot appear in the same group of 4 pairs.

So {A-B, C-D, E-F, G-H} is one set

{A-C, B-D, E-F, G-H} is another set... and perhaps there are 103 more such sets, but how to mathematically arrive at it?
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by sankruth » Wed Jan 23, 2008 8:38 am
Oh...the solution briefly explains how but I cant understand it...

Here is the explanation...
out of 8 people one team can be formed in 8c2 ways.

8c2*6c2*4c2*2c2= 2520.
The answer is 105. Divide 2520 by 4! to remove the multiples ( for example: (A,B) is same as ( B,A) )

But for Combinations A-B is the same as B-A and isnt counted twice, so the above explanation seems wrong to me.
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by Stuart@KaplanGMAT » Wed Jan 23, 2008 12:06 pm
I solved a different way. Let's call our players A, B, C, D, E, F, G and H.

For the first pair, we have 8 choices for slot 1 and 7 choices for slot 2. For the second pair, we have 6 choices for slot 1 and 5 for slot 2. For the 3rd pair we have 4 choices for slot 1 and 3 choices for slot 2. For the 4th pair we have 2 choices for slot 1 and 1 for slot 2.

So, so far we have:

(8*7) (6*5) (4*3) (2*1)

Next, we have to remember that we don't care about the order in which we choose the members of each pair (since, e.g. AB is the same as BA) - so we need to divide each pair by 2 to eliminate those duplicates.

Now we have:

8!/2*2*2*2 = 2520, the answer at which everyone has arrived so far.

However, we're not done yet.

We need to recognize that our pairings could have been:

(AB) (CD) (EF) (GH)

or

(GH) (CD) (EF) (AB)

and that both of those results are actually identical teams of 2. We have more duplicates to eliminate.

Since 4 terms can be arranged in 4! different ways, we need to divide our final result by 4! to eliminate all of the duplicate arrangements.

Therefore, our final answer is:

8!/2*2*2*2*4! = 8*7*6*5*4!/16*4! = 8*7*6*5/16 = 7*3*5 = 105

Choose (E)
Last edited by Stuart@KaplanGMAT on Wed Jan 23, 2008 1:36 pm, edited 1 time in total.
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by sankruth » Wed Jan 23, 2008 1:14 pm
Stuart, you are a star!!

Thank you very much!
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by cris » Wed Jan 23, 2008 4:38 pm
Wow, Stuart that was an amazing approach!
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by simplyjat » Thu Jan 24, 2008 2:53 am
sankruth wrote:Oh...the solution briefly explains how but I cant understand it...

Here is the explanation...
out of 8 people one team can be formed in 8c2 ways.

8c2*6c2*4c2*2c2= 2520.
The answer is 105. Divide 2520 by 4! to remove the multiples ( for example: (A,B) is same as ( B,A) )

But for Combinations A-B is the same as B-A and isnt counted twice, so the above explanation seems wrong to me.
You have to still divide by 4! to get the right answer, but the reason is different.

The ordering of doubles do not result in different teams.... i.e. T1, T2, T3, T4 is same as T4, T2, T3, T1 and T1, T3, T2, T4..... and so on.... As there are four teams we end in 4P4 ways of arranging the teams, and all these are same set of teams.... Thus the answer has to be....

8C2*6C4*4C2*2C2/4P4....
simplyjat
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by viidyasagar » Sat Jan 30, 2010 12:41 am
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
Although Stuart's approach is logical, i have come across a lot of people struggling to move beyond 2520.

Here's a thumb rule approach and IMHO, permutations&C and probability can be learnt mechanically as well.

In general, The number of ways in which m*n different items can be divided equally into m groups, each containing n objects and the order of the groups is not important is

[(mn)! / ((n!)^m)* m!)]

Here, we have 4*2 people, who have to be divided into 4 groups, each group containing 2 people and the order in not important, then

A. Numerator = (4*2)! = 8!
B. Denominator = (2!^4)*4!

A/B = 105.

If order of the groups is important then do not divide the expression by m! and u are done

Hope this helps!!!
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