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m06Q11

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Author Message
Senior Manager
Senior Manager
Status: Student
Joined: 26 Aug 2013
Posts: 262
Location: France
Concentration: Finance, General Management
GMAT 1: 650 Q47 V32
GPA: 3.44
Followers: 2

Kudos [?]: 25 [0], given: 397

Re: m06Q11 [#permalink] New post 28 Oct 2013, 05:26
Hi all!

I personally understand it like this:

You have (7*4)*(3*5)*(3*2)*1 combinations. (If you do not understand there are some really good posts that are explaining it really well, espacially the one of Brunel)

BUT the questions says that we need to find the number of different teams.

For the first one, the total of the possibilities is 7*4. That means that you have 7 possibilities, and that the team can be spread in 4 groups! But here, we are only looking for the number of different teams. Therefore, if groupe AB can be spread in team 1, 2, 3 or 4 = the number of different combinations is 4 but we still have ONE same team.

In consequence, what do we have to do? Divide by 4 for (7*4), 3 for (3*5), 2 for (2*3) or in other term divide the whole by 4!

Answer B.

Hope it helps.
_________________

Think outside the box

Re: m06Q11   [#permalink] 28 Oct 2013, 05:26
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