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# Ben and Ann are among 7 contestants from which 4

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SVP
Joined: 03 Feb 2003
Posts: 1604
Ben and Ann are among 7 contestants from which 4 [#permalink]

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04 Jul 2004, 23:32
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Ben and Ann are among 7 contestants from which 4 semifinalists are to be selected. Of the different possible selections, how many contain both Ben and Ann?
Senior Manager
Joined: 21 Mar 2004
Posts: 445
Location: Cary,NC

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05 Jul 2004, 00:01
no of combinations containing B and A = 5C2 ??
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ash
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I'm crossing the bridge.........

Manager
Joined: 02 Jun 2004
Posts: 184
Location: Kiev, Ukraine

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05 Jul 2004, 00:40
stolyar wrote:
Ben and Ann are among 7 contestants from which 4 semifinalists are to be selected. Of the different possible selections, how many contain both Ben and Ann?

stolyar, don't you think that such questions follow simple pattern? I see them again and again here...

OK, C[2,5] = 10 is the answer.
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SVP
Joined: 03 Feb 2003
Posts: 1604

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05 Jul 2004, 00:50
I see what you, a veteran, mean. The total number of registered users is to reach 6000 soon. There are many many new visitors who look through the most recent posts. Let them enjoy what you already know.
Joined: 31 Dec 1969
Location: Russian Federation
GMAT 3: 740 Q40 V50
GMAT 4: 700 Q48 V38
GMAT 5: 710 Q45 V41
GMAT 6: 680 Q47 V36
GMAT 9: 740 Q49 V42
GMAT 11: 500 Q47 V33
GMAT 14: 760 Q49 V44
WE: Supply Chain Management (Energy and Utilities)

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07 Jul 2004, 12:14
Can someone please explain the simple rationale behind this? I understand the "2" part of the combination, but why "5"?

Thanks!
SVP
Joined: 03 Feb 2003
Posts: 1604

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08 Jul 2004, 04:53
5 are people who are neither Ben nor Ann. Among them we select 2.
Intern
Joined: 09 May 2004
Posts: 27
Location: houston, tx

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08 Jul 2004, 10:24
There are four places, 7 people to choose from. Ben and Ann have to be selected, so that leaves two places left out of the four that can be filled. There are 5 remaining contestants to choose from that can fill those two spaces. Therefore 5C2.

Combinations involving Ben and Ann with two other people =
5! / (2! 3!) = 10 combos
Manager
Joined: 16 Jan 2004
Posts: 64
Location: NJ

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09 Jul 2004, 14:06
Why cant it be something like this

total ways in which A &B together

=total ways - no of ways in which they are not together
= 7C4 - 5C4 (If they are not together we have total 5 and choose 4

Obviously this is wrong BUT WHY?
09 Jul 2004, 14:06
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