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

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

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

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

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

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

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

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.

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

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.

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

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.

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 18C2- (6x3)
Ans 135

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


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

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

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

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