There are 10 people in a room. If each person shakes hands with exactl

why do we have to divide 30 by 2...
substituting 5 with 10, i would get 5 which is half of 10.

Hello Karishma

Can you please explain further?
I think I don't quite understand it.

Is is so, that when you want to shake hand with 3 people, the number of persons in the group has to be even and equal or greater than 4?

Thanks

Consider you and I shake hands. When I count 30, I have counted it twice. Once for you, and once for me. But it is actually just one handshake. Each one of those 30 events was one hand that was shaken. Two of those events make one handshake so we divide by 2.

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]

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.

Since it is given that each person does shake hands with exactly 3 people, it will certainly be possible. Here is one such instance:

1&2
3&4
5&6
7&8
9&10
(Each person shakes hands once)

1&3
2&5
4&7
6&9
8&10
(Each person shakes hand with another person)

1&5
2&9
3&7
4&10
6&8
(Each person shakes hands with yet another person)

Can you please suggest more questions like this one to practice?
I struggled with this one really..my approach was different for solving this. Your approach is really creative, and without having solved something of this sort before or read about the method you used, I wouldn't have been able to do this question. Please also suggest something to read.
Thank you so much

if each of the 10 had shaken hands once with 9 others,
there would have been 10*9/2, or 45 total handshakes,
but, because only one third (3 out of 9) of possible handshakes take place,
there are only one third (15 out of 45) total handshakes
15
A

Hi All,

Since the answer choices are so "spread out", there's an interesting way to get to the correct answer by avoiding complex math and using "brute force" and a comparison:

We're told that there are 10 people in the room and that each person shakes hands with 3 other people.

Let's say there were 4 people, who we'll call A, B, C and D.

The handshakes would be:
AB
AC
AD
BC
BD
CD

In this situation, each person shook hands with 3 people and there was a total of 6 handshakes.

If we TRIPLED the number of people, then we'd have 12 people, and we'd have TRIPLE the handshakes: 6 x 3 = 18.

Since we have FEWER than 12 people, we'll have FEWER than 18 handshakes. There's only one answer that fits:

Final Answer:

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

Here's an approach that doesn't require any counting techniques:

Each person shakes hands with exactly 3 other people
So, we have 10 people and each experiences 3 handshakes for a TOTAL of 30 handshakes.

IMPORTANT: at this point, we need to recognize that every handshake has been counted TWICE. For example, if Person A and Person B shake hands, then Person A counts it as a handshake, AND Person B also counts it as a handshake. Of course only one handshake occurred.

To account for the DUPLICATION, we'll divide 30 by 2 to get 15

Answer: A

Cheers,
Brent

VeritasKarishma,

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

Will there be only 6 handshakes ?

A: B, C, D -->3
B: C, D -->2
C: D --1
D:
E =0 because A, B, C, and D already shake hands with each other and only 3 exactly handshakes are allowed. So no one is left to shake hand with E.

Is my thinking correct ?

Note this:
AB, CD,
EA, BC,
AC, BE,
DE

A, B, C and E have 3 handshakes each but D has only 2 handshakes.
It is not possible for D to have a 3rd handshake. That is why we get 7.5 when we use the formal method.

VeritasKarishma ,Chetan sir,

I need your help in this!!

I understood the explanation that you gave but this reminded me of an OG problem

There are 8 teams in a certain league and each team plays each of the other teams exactly once. If each game is played by 2 teams, what is the total number of games played?

A. 15
B. 16
C. 28
D. 56
E. 64

This problem I understand we have to make selection of 2 players out of 10..

So I was little confused ..isn't it similar to the handsake problem??

If Team A player say 8th player plays with the 7th player --that means 7th has played with 8th..(much like handsake problem)

The only difference is in handsake is done by 3 people and here game is played by two teams..

If I use that approach..which I am very tempted to..I land up with a wrong answer..

Can you please help me here?

Hi prabsahi,

The question that you're asking about is discussed here:

https://gmatclub.com/forum/there-are-8- ... 34582.html

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

Yes, the problems are similar, though not same. So you need to tweak the numbers a bit.

In the games problem, every team plays with "all" other teams (so each team plays with 7 other teams). In every game, exactly 2 teams are involved.
No of games = 8*7/2 = 28

In the handshake problem, every person shakes hands with 3 other people. In every handshake, exactly 2 people participate.
No of handshakes = 10*3/2 = 15

Say, there are 4 people A,B,C,D: each makes exactly 2 other handshakes,
Possibility: AB AC, BC BD, CD AD, AD BD
Overlapping handshakes: AD BD = 2
Unique handshakes: AB AC BC CD= 4
Therefore, total handshakes= 4+2=6

Per the logic used above of halving the total handshakes: 4*2/2, the answer should be 4. It does not always hold true.

It really depends on the percentage of overlapping handshakes. In the scenario where each of ABCD shakes the hand of every other individual, there is clearly an overlap or double counting of all the handshakes.

But in our case, where people are selectively shaking the hands of not all other people but few (in this example, 2) other people, the uniques handshakes will depend on the percentage overlap.

Can someone give a more solid approach to this?

Look at your example again.

C participates in 3 handshakes, not the 2 you specify.

Pairs that work are for example:

AB BC AD and CD or

AB AC CD BD

but in each example only 4 pairs occur, not 6.

Posted from my mobile device