VeritasPrepKarishma wrote:

shrive555 wrote:

There are 10 people in a room. If each person shakes hands with exactly 3 other people, what is the total number of handshakes?

15

30

45

60

120

Consider one hand that gets shaken as one event. One person shakes his hand with three other people so 3 events take place per person. In all 10*3 = 30 events take place. But two hands make one handshake. So 30 of these events will make 15 handshakes.

Question: Substitute 5 in place of 10. What do you get? Why?

[/quote]

Ok Look.

4 people: shrive555, craky, karishma and Mr X

We have to shake hands in this group such that each person shakes hands with 3 people.

So shrive555 starts:

shrive555 - craky : 1 handshake but 2 hands were shaken. shrive555's and craky's

shrive555 - karishma : 1 handshake but 2 hands were shaken. shrive555's and karishma's

shrive555 - Mr X : 1 handshake but 2 hands were shaken. shrive555's and Mr X's

Now shrive555 has shaken hands with 3 people. There were 3 handshakes. But 6 hands were shaken.

Now, when all 4 of us shake hands with 3 people, each person's hand will be shaken 3 times. i.e. in all 12 hands will be shaken. But they will add up to only 6 handshakes.

The other 3 handshakes will be:

craky - karishma

craky - Mr X

karishma - Mr X

So all of us have shaken hands with exactly 3 people.

Similarly when there are

10 people, each person shakes his hand 3 times. So in all 30. But 2 of these hands combined to make one handshake. So we will get only 15 total handshakes.

Now when there are

5 people and each person has to shake hands with exactly 3 people, each person will shake his hand 3 times. We will have total 15 hands that will be shaken. So how many handshakes does it make? 7.5? That is not possible. This is because it is not possible for 5 people to shake hands such that each person will shake hands with exactly 3 people. Can you think of the condition which must be satisfied such that it is possible that each person shakes hands with exactly 3 people? (craky, you are there. Just think some more to be clear.)[/quote]

Hi Karishma,

Take this forward for 10 individuals we have the following scenario:

1 & 2, - 1st unique handshake

1 & 3, - 2nd unique handshake

1 & 4, - 3rd unique handshake

2 & 1,

2 & 3, - 4th unique handshake

2 & 4, - 5th unique handshake

3 & 1,

3 & 2,

3 & 4, - 6th unique handshake

Case for individual number 5, 6, 7 and 8 will also make 6 unique handshakes. So in all 12 upto now.

For individual number 9 and 10.

9 & 10 - 11th unique handshake

10 & 9

Now, individuals numbered 9 and 10 don't even end up meeting the condition of "exactly 3 handshakes".

I selected 15, as it was the closest (to 11 unique handshakes) choice. Can you please assist? Thanks.