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by radhika1306 » Mon Sep 17, 2007 6:52 pm
In a meeting of 3 representatives from each of 6 different companies, each person shook hands
with every person not from his or her own company. If the representatives did not shake hands
with people from their own company, how many handshakes took place?
A. 45
B. 135
C. 144
D. 270
E. 288
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by kajcha » Mon Sep 17, 2007 7:05 pm
Ans D

There are 18 people in conf and each shaked hand with 15 people

So 18*15 = 270
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by niks_01.27 » Mon Sep 17, 2007 7:06 pm
There are total 18 person. Each person shook hands 15 persons (leaving 2 from his own company and 1, himself).

So total nbr of hand shakes are 15 x 18 = 270.

Whats the answer?
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by niks_01.27 » Mon Sep 17, 2007 7:28 pm
Anonymous wrote:I am getting (B) as the answer.

Team 1 shakes hand with 15 people and hence 3x15 =45
Team 2 has already shaked hands with Team 1.
Team 2 shakes hand with 12 people and hence 3x12 =36
and so on gives,

3x15+3x12+3x9+3x6+3x3=45+36+27+18+9=135.
My bad, I was in a hurry to solve the question. Your answer makes sense.
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by radhika1306 » Tue Sep 18, 2007 7:39 am
Answer is B
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by apple100 » Sun Mar 22, 2009 3:17 pm
Anonymous wrote:I am getting (B) as the answer.

Team 1 shakes hand with 15 people and hence 3x15 =45
Team 2 has already shaked hands with Team 1.
Team 2 shakes hand with 12 people and hence 3x12 =36
and so on gives,

3x15+3x12+3x9+3x6+3x3=45+36+27+18+9=135.
is there a shortcut to this problem besides this method?
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by lilu » Sun Mar 22, 2009 3:53 pm
Each person has to shake hands with 15 people (18 total-2 from his team -1 himself=15)
And there are 18 people *15 handshakes = 270,
now we need 2 people to make a handshake, so we need to divide 270 by 2-->135, otherwise we'd be counting each combination two times
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