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# GMAT Question-Combinatorics

Author Message
Senior Manager
Joined: 18 Aug 2009
Posts: 425
Schools: UT at Austin, Indiana State University, UC at Berkeley
WE 1: 5.5
WE 2: 5.5
WE 3: 6.0
Followers: 8

Kudos [?]: 129 [0], given: 16

### Show Tags

22 Sep 2009, 18:25
00:00

Difficulty:

(N/A)

Question Stats:

100% (01:14) correct 0% (00:00) wrong based on 2 sessions

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This topic is locked. If you want to discuss this question please re-post it in the respective forum.

How many different handshakes are possible if six girls are standing on a circle and each girl shakes hands with every other girl except the two girls standing next to her?

A)12
B)11
C)10
D)9
E)8
_________________

Never give up,,,

Manager
Joined: 11 Sep 2009
Posts: 129
Followers: 6

Kudos [?]: 362 [1] , given: 6

### Show Tags

22 Sep 2009, 18:46
1
KUDOS

Basically, every girl is able to shake hands with 3 other people (6 people - 2 people beside them - themselves). So you have 6 people * 3 handshakes/person. Each handshake, however, is going to be counted twice, since there's one person on each end of the handshake. So the total amount of handshakes possible is:

6 * 3 / 2
= 9
Senior Manager
Joined: 31 Aug 2009
Posts: 417
Location: Sydney, Australia
Followers: 9

Kudos [?]: 295 [0], given: 20

### Show Tags

22 Sep 2009, 19:19

Because it looked like a manageable number (6 girls) I thought brute force would be quick…
1 -> 3,4,5
2 -> 4,5,6
3 -> 1,5,6 (1 Double count)
4 -> 1,2,6 (2 Double count)
5 -> 1,2,3 (3 Double count)
6 -> 2,3,4 (3 Double count)
My working on paper made more sense and didn't take long and I just striked-out double counts took me around 1:00

I’m not good at perm/combs but I redid it with that anyway:
Question can be paraphrased to be how many pairs of girls except 6 combos: 6C2 = 6! / 2!4! = 15
15 combos in total
6 combos disallowed (cannot shake hands with girl on left covers the clause)
Total combos: 15-6 = 9
Re: GMAT Question-Combinatorics   [#permalink] 22 Sep 2009, 19:19
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