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doubt: combinations question

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Manager
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Joined: 11 Apr 2011
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doubt: combinations question [#permalink] New post 14 Jul 2011, 05:40
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A
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D
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(N/A)

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0% (00:00) correct 0% (00:00) wrong based on 0 sessions
There are 4 pairs of siblings. In how many ways 4 teams for a tennis match can be formed so that none of the siblings are in the same team?
Manager
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Re: doubt: combinations question [#permalink] New post 14 Jul 2011, 08:52
i don't know the answer. I just thought about the question while doing a similar question. I was not able to crack this question..
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Re: doubt: combinations question [#permalink] New post 14 Jul 2011, 13:06
i think the answer is 60.. phew!! tough one.. i dont know whether my approach is correct.. C&C welcome..

starting from ground up .. consider the simplest scenario
---------
2 pairs of siblings - (ab) (cd)

possible team combinations:
a c + b d &
a d + b c

the way I look at it is .. a[c/ d] = two ways
makes sense.. if it doesnt .. pls stop reading here.. rest will all be gibberish..
--------

3 pairs of siblings - (ab) (cd) (ef )
possible combinations

a [c/ d/ e/ f] = 4 ways

remaining players are always similar to (b) (d) (ef) which can be arranged in 2 ways in turn

=> two players without their siblings and one sibling pair

so total = 8 ways

----------

4 pairs of siblings (ab) (cd) (ef) (gh)

a [c/ d/ e/ f/ g/ h] = 6 ways and remainng will be similar to

(b) (d) (ef) (gh)

I can arrange them by picking the first two (bd) and rearrange the remaining 4 into two teams = 2 ways
or I can pick one single sibling and other from sibling pair like b [e/ f/ g/ h] and with d in two ways = 8 ways

so total ways = 6 * (2 + 8 ) = 60 ways..
Manager
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Joined: 11 Apr 2011
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Re: doubt: combinations question [#permalink] New post 18 Jul 2011, 10:33
anyone kindly address this question.. i m still struck on this..
Re: doubt: combinations question   [#permalink] 18 Jul 2011, 10:33
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