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# Hand shakes

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Senior Manager
Joined: 13 Aug 2010
Posts: 314
Followers: 1

Kudos [?]: 13 [0], given: 1

Hand shakes [#permalink]  12 Oct 2010, 20:27
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Question Stats:

100% (00:00) correct 0% (00:00) wrong based on 2 sessions
Everyone shakes hands with everyone else in a room. Total number of handshakes is 66. Number of persons=?

a.14
b.12
c.11
d.15
e.16
[Reveal] Spoiler: OA
Retired Moderator
Joined: 02 Sep 2010
Posts: 805
Location: London
Followers: 87

Kudos [?]: 639 [0], given: 25

Re: Hand shakes [#permalink]  12 Oct 2010, 21:06
prab wrote:
Everyone shakes hands with everyone else in a room. Total number of handshakes is 66. Number of persons=?

a.14
b.12
c.11
d.15
e.16

In a room of n people, the number of possible handshakes is C(n,2) or n(n-1)/2

So n(n-1)/2 = 66 OR n(n-1)=132 OR n=12

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Senior Manager
Joined: 13 Aug 2010
Posts: 314
Followers: 1

Kudos [?]: 13 [0], given: 1

Re: Hand shakes [#permalink]  12 Oct 2010, 21:19
can you please explain why are we using n(n-1)/2, m not able to grab the concept.
Retired Moderator
Joined: 02 Sep 2010
Posts: 805
Location: London
Followers: 87

Kudos [?]: 639 [0], given: 25

Re: Hand shakes [#permalink]  12 Oct 2010, 21:34
Number of handshakes will be the number of ways to choose 2 people out of n people. For every choice of two people, there is a handshake.

This number is C(n,2) = $$\frac{n!}{(n-2)!2!}$$
_________________
Re: Hand shakes   [#permalink] 12 Oct 2010, 21:34
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