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3 representative from 6 companies. shake hands only with

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CEO
Joined: 15 Aug 2003
Posts: 3467
Followers: 61

Kudos [?]: 718 [0], given: 781

3 representative from 6 companies. shake hands only with [#permalink]  13 Sep 2003, 05:38
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3 representative from 6 companies. shake hands only with other
companies' representative. representative from same company does not
shake hands. how many hand shakes?

Check my work

Total # of handshakes = 18 * 17 = 306

We double count the handshakes and also count those amongst representatives ...

Dividing by 2 ...So total handshakes = 306/2 =153

For Handshakes amongst representatives...
Three people => Handshakes= 3*2/2 = 3
For six companies = > total handshakes = 6 * 3 =18

153 - 18 = 135 ..Is this correct?

thanks
Senior Manager
Joined: 22 May 2003
Posts: 334
Location: Uruguay
Followers: 1

Kudos [?]: 41 [0], given: 0

[#permalink]  13 Sep 2003, 20:45
I guess so. I got the same answer.

I did 3*(3*5+3*4+3*3+2*3+3)=3*(15+12+9+6+3)=135
Intern
Joined: 13 Sep 2003
Posts: 43
Location: US
Followers: 0

Kudos [?]: 8 [0], given: 0

[#permalink]  04 Nov 2003, 08:45
another way to look at the same problem:

for any handshake u need 2 persons.

so total number of handshakes between 18 ppl = 18C2 = 153 .
Now within the group ; 3C2 = 3.
Total groups 6 so 6*3 = 18.

Now total valid handshakes for this scenario is 153- 18 = 135.
Intern
Joined: 13 Sep 2003
Posts: 43
Location: US
Followers: 0

Kudos [?]: 8 [0], given: 0

formula [#permalink]  10 Nov 2003, 09:20
1.Number of handshakes between n people is n(n-1)/2.

2.Number of diagnols of a polygon with n sides = n(n-1)/2 - n.
formula   [#permalink] 10 Nov 2003, 09:20
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3 representative from 6 companies. shake hands only with

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