Handshake Problem - Six companies and three employees

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Hey, may anyone help me with the problem bellow?
thanks!

Six companies attend a workshop - each is represented by three employees. All employees, except those from the same companies, shake hands to greet. How many hand-shakes were exchanged?

a) 102 b) 123 c) 135 d) 193 e) 288

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by GMATGuruNY » Mon Jul 11, 2016 5:19 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
Total number of representatives = (number of representatives per company)(total number of companies) = 3*6 = 18.

Good handshakes = total possible handshakes - bad handshakes.

Total possible handshakes:
From the 18 representatives, the total number of pairs that could shake hands = 18C2 = (18*17)/(2*1) = 153.

Bad handshakes:
A bad handshake occurs when a pair of representatives from the same company shake hands.
From each group of 3 representatives from the same company, the total number of pairs that could shake hands = 3C2 = (3*2)/(2*1) = 3.
Since there are 6 companies -- and each company yields a total of 3 bad handshakes -- the total number of bad handshakes = 6*3 = 18.

Good handshakes:
(total possible handshakes) - (bad handshakes) = 153 - 18 = 135.

The correct answer is B.
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by [email protected] » Mon Jul 11, 2016 5:26 pm
Hi henriqueargollo,

These types of question can often be solved with some basic note-taking and some 'brute force' math.

I'm going to refer to the employees as...

AAA
BBB
CCC
DDD
EEE
FFF

Each of the "A" employees will shake hands with each of the BCDEF employees, giving us (3)(15) = 45 handshakes.

Since each of the "B" employees have ALREADY shaken hands with the "A" employees, they'll then shake hands with the CDEF employees. This gives us (3)(12) = 36 additional handshakes.

In that same way, the "C"s shake hands with each of the DEF employees, giving us (3)(9) = 27 more handshakes

The "D"s shake hands with each of the EF employees, giving us (3)(6) = 18 more handshakes
And the "E"s shake hands with each of the F employees, giving us (3)(3) = 9 more handshakes

Total handshakes = 45 + 36 + 27 + 18 + 9 = 135 handshakes

Final Answer: C

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by OptimusPrep » Mon Jul 11, 2016 8:20 pm
henriqueargollo wrote:Hey, may anyone help me with the problem bellow?
thanks!

Six companies attend a workshop - each is represented by three employees. All employees, except those from the same companies, shake hands to greet. How many hand-shakes were exchanged?

a) 102 b) 123 c) 135 d) 193 e) 288
Total employees = 6*3 = 18
Total number of handshakes = 18C2 = 153

Undesired handshakes = hand shakes between employers of same company = 3C2 = 3
Total undesired handshakes = 6*3 = 18

Total handshakes exchanged = 153 - 18 = 135

Correct Option: C

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by Brent@GMATPrepNow » Tue Jul 12, 2016 5:48 am
henriqueargollo wrote:
Six companies attend a workshop - each is represented by three employees. All employees, except those from the same companies, shake hands to greet. How many hand-shakes were exchanged?

a) 102 b) 123 c) 135 d) 193 e) 288
There are 18 attendees altogether.

If you ask any attendee, "How many people did you shake hands with?", he/she will say 15, because that person will shake hands with the 15 people that are not in his/her company.

So, the TOTAL number of handshakes = (18)(15) = 270

But wait, we have counted every handshake TWICE. Person A counted his/her handshake with person G, and person G counted his/her handshake with person person A.

To account for this duplication, we must divide 270 by 2 to get 135.

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by Matt@VeritasPrep » Wed Jul 20, 2016 9:08 pm
I'd think of it as

Total - Same Company

Total = (18 choose 2), since we've got 18 people and we're choosing 2 for each handshake.

Same Company = 6 companies * (3 choose 2), since each company has 3 people, and we'd be choosing two for each (unwanted!) handshake.

This gives us

(18 choose 2) - 6 * (3 choose 2) =>

18!/(16! * 2!) - 6 * 3!/(2! * 1!) =>

(18*17 / 2) - 6 * 3 => 153 - 18 => 135