In a meeting of 3 representatives from each of 6 different companies,

Question

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

A. 45
B. 135
C. 144
D. 270
E. 288

chetan2u · Accepted Answer

Another way

Lets count the total number of handshakes = 18C2= 153..

Out of these within company handshakes = 3C2=3
and no of companies=6
so total handshakes within company = 6*3=18

Handshakes with given restrictions = 153-18= 135 
B

BrentGMATPrepNow · Answer

Let's focus on 1 person, call him Ted from company A. 
Ted will shake hands with a total of  15  people (all 3 people who are in the other 5 companies).
Likewise, Ann from Company B will also shake hands with  15  people. 
And so on....

In fact, all  18  people will shake hands with  15  others.

So, it SEEMS like the TOTAL number of handshakes = ( 18 )( 15 )
HOWEVER, we need to keep in mind that we have counted each handshake TWICE. 
That is, if Ted shakes hands with Ann, then we have counted that handshake once in Ted's  15  handshakes, AND once in Ann's  15  handshakes.
And so on...

To account for this DUPLICATION, we must divide ( 18 )( 15 ) by 2. 
So, the TOTAL # of handshakes = ( 18 )( 15 )/2 = 135 = B

Cheers,
Brent

prashant212 · Answer

I don't know any shortcut to this question but here's what i counted:

There are total 18 people from 6 companies,3 from each.Now each of them starts shaking hands one by one with other companies representative. If the companies are A,B,C,D,E,F.

Each member from company A gets to shake hands with 15 people(Total 18 minus member from his company and himself). 
Then each member from company B shakes hands with 12 people because they already shook hands with company A members. Similarly company C member shakes hand with 9 members and so on.

so we have (15*3)+(12*3)+(9*3)+(6*3)+(3*3) =  135

chetan2u · Answer

Hi  prashant212 ,
 a short cut would be... 
 choose two companies out of 6 = 6C2..
the hand shakes with in these two companies = 3*3=9..
Total handshakes = 6C2 * 9 = 15 * 9 =135 
B

OptimusPrepJanielle · Answer

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: B

mikeyg51 · Answer

Excellent explanation and I only got this right going the long route....I am struggling with combinations more than I thought I would...I get a few steps correct but then leave a crucial one out. I will continue to practice using your laid out approach

SVaidyaraman · Answer

1. It is a combination problem
2. Is there a constraint? The constraint is a person does not shake hand with another person of his own company.
3. Assume the opposite of the constraint. The person shakes hand with another person of the same company
4. Total number of combinations is 18C2= 153
5. Opposite of the constraint is 3 handshakes within same company men * total of 6 companies=18 
6. Number of handshakes with constraints is (4)-(5)= 135

JeffTargetTestPrep · Answer

We are given that there are 3 representatives from 6 different companies. So, there are a total of 18 representatives.
 
If every representative were to shake hands with all other representatives (meaning all 18 reps would shake hands), this would happen in the following number of ways:
 
18C2 = (18 x 17)/2! = 9 x 17 = 153 ways
 
However, since each person shook hands with every person not from his or her own company, we can subtract out the number of times those handshakes occurred. 
 
Since each company has 3 reps, the number ways those three reps can shake hand is 3C2 = (3 x 2)/2! = 3 ways, and since there are 6 companies, this would occur 6 x 3 = 18 times.
 
Thus, the number of ways for the reps to shake hands with every person not from his or her own company is 153 - 18 = 135 ways.
 
Answer: B

SISDIT · Answer

Hi all,

Each company has 3 representatives and there are 6 companies. So there are 18 people in the meeting.

One handshake needs 2 people participating since people are NOT shaking their own hands.

So the question would be: in how many ways we can choose 2 people out of 18 people.

Let say there are 2 slots: __A __B

There are 18 possibilities for A.

There are 15 possibilities for B. Why?

+The person picked for A can NOT be shaking his/her own hand. So there are 18 - 1 = 17 people left to choose for B

+The person picked for A also can NOT be shaking his/her colleagues' hands. So there are 17 - 2 = 15 people left to choose for B.

So there are 18 x 15 = 270 handshakes, assuming that AB is DIFFERENT from BA. This means that A shaking B's hand is DIFFERENT from B shaking A's hands.

But, logically they are all the SAME -- A shakes B's hand = B shakes A's hand.

So of 270 handshakes, there are 2! handshakes being OVERCOUNTED.

So 270 needs to be divided by 2! to eliminate overcounting handshakes.

Answer = 270 / 2! = 135.

I am not sure about my approach. Could any one shed some light on that. Really appreciate.

EMPOWERgmatRichC · Answer

Hi All,

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:  B

GMAT assassins aren't born, they're made, 
Rich

US09 · Answer

I don't know if my approach is right:

Total number of reps: 18
Total no of reps to shake hands with (excluding his/her own company reps and obviously his/her ownself): 15
Therefore, total number of handshakes: 18*15=270
But handshake already involves two reps, i.e. when A shook hands with B, B also shook hands with A.
Thus, effective total: 270/2=135.

B.

traxx1 · Answer

Select 2 out of 18= 153
Select 2 out of 3 = 3
Select 2 out of 3 for all 6 companies = 3* 6 = 18
153- 18 = 135

mgmat4944 · Answer

There are 18 people:

Organize these 18 people into 6 groups of 3 ( each group of 3 is from the same company)

Then you chose 2 groups from the 6 groups. (6C2)

Then from the 2 groups there are 3 ways to pick pairs. Have to remember that Person A, Person B handshake is the same as Person B Person A handshake. so it is 3*6C2 = 135.

Businessconquerer · Answer

SUCH QUESTIONS JUST DEPEND ON STRIKING
Lets look at this one the easy way

All shook hands - the people who didn't will give you the result.
That is 18 C 2- (6x3)
Ans 135

Archit3110 · Answer

Total handshakes 18c2
Restriction 3c2×6:18
18c2-18; 135
IMO B

Posted from my mobile device

abalodi · Answer

can someone please explain why the total # of handshakes is 18C2. Why are we choosing 2 handshakes? What is the logic via this approach?

chetan2u · Answer

Since any handshake requires two people, we are counting different ways 2 people can be selected out of 18 people.

abalodi · Answer

That seems all to obvious but now it all makes sense. Appreciate the help!

Posted from my mobile device

Kritisood · Answer

Could someone explain why 15c2 is wrong? which scenario am I missing in this?

In a meeting of 3 representatives from each of 6 different companies,

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