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Everyone shakes hands with everyone else in a room. Total number of ha

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Senior Manager
Joined: 13 Aug 2010
Posts: 281

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Everyone shakes hands with everyone else in a room. Total number of ha [#permalink]

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12 Oct 2010, 21:27
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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

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Joined: 02 Sep 2010
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Re: Everyone shakes hands with everyone else in a room. Total number of ha [#permalink]

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12 Oct 2010, 22:06
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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Kudos [?]: 1184 [0], given: 25

Senior Manager
Joined: 13 Aug 2010
Posts: 281

Kudos [?]: 35 [0], given: 1

Re: Everyone shakes hands with everyone else in a room. Total number of ha [#permalink]

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12 Oct 2010, 22:19
can you please explain why are we using n(n-1)/2, m not able to grab the concept.

Kudos [?]: 35 [0], given: 1

Retired Moderator
Joined: 02 Sep 2010
Posts: 793

Kudos [?]: 1184 [0], given: 25

Location: London
Re: Everyone shakes hands with everyone else in a room. Total number of ha [#permalink]

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12 Oct 2010, 22:34
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) = $$\frac{n!}{(n-2)!2!}$$
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Kudos [?]: 1184 [0], given: 25

Re: Everyone shakes hands with everyone else in a room. Total number of ha   [#permalink] 12 Oct 2010, 22:34
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