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by Brent@GMATPrepNow » Wed Aug 05, 2015 11:38 am
There are 10 people in a room. If each person shakes hands with exactly 3 other people, what is the total number of handshakes?

A) 15
B) 30
C) 45
D) 60
E) 120
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 person 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

Here's a similar question: https://www.beatthegmat.com/count-questions-t274369.html

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by GMATGuruNY » Wed Aug 05, 2015 11:41 am
There are 10 people in a room. If each person shakes hands with exactly 3 other people, what is the total number of handshakes?

A) 15
B) 30
C) 45
D) 60
E) 120
Alternate approach:

The total number of pairs that can be formed from 10 people = 10C2 = 45.
If all of these pairs shake hands, each person in the room will shake hands with EVERY OTHER PERSON in the room.
The result will be 9 HANDSHAKES PER PERSON.
Since we want only 3 HANDSHAKES PER PERSON, we must divide by 3:
45/3 = 15.

The correct answer is A.
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by nikhilgmat31 » Tue Aug 11, 2015 9:32 pm
we can solve it simply as

if 1 person shake hand with 3 persons - total Hand shakes will be 3

simply 10 person will shake it 10 * 3 = 30 times.

But some the persons have already shaked hands so it must be less than 30.

there is only one answer option less than 30 - 15 is answer. :)
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by [email protected] » Sun Aug 30, 2015 11:28 am
GMATGuruNY wrote:

The total number of pairs that can be formed from 10 people = 10C2 = 45.
If all of these pairs shake hands, each person in the room will shake hands with EVERY OTHER PERSON in the room.
The result will be 9 HANDSHAKES PER PERSON.

/quote]

Hi Mick

How do 45 pairs contribute to 9 handshakes/person.
Please elaborate.

Thanks :)
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by [email protected] » Sun Aug 30, 2015 2:18 pm
Hi ahmedbari.ace,

Here is how you can 'visualize' 10 people each shaking hands with one another.

Let's name the people A,B,C,D,E F,G,H,I,J

The 45 handshakes would be....

AB,AC,AD,AE,AF,AG,AH,AI,AJ
BC,BD,BE,BF,BG,BH,BI,BJ
CD,CE,CF,CG,CH,CI,CJ
DE,DF,DG,DH,DI,DJ
EF,EG,EH,EI,EJ
FG,FH,FI,FJ
GH,GI,GJ
HI,HJ
IJ

You'll notice that there are 9+8+7+6+5+4+3+2+1 = 45 handshakes and that each letter appears in exactly 9 different handshakes.

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by MartyMurray » Sun Aug 30, 2015 8:03 pm
[email protected] wrote:
GMATGuruNY wrote:

The total number of pairs that can be formed from 10 people = 10C2 = 45.
If all of these pairs shake hands, each person in the room will shake hands with EVERY OTHER PERSON in the room.
The result will be 9 HANDSHAKES PER PERSON.

/quote]

Hi Mick

How do 45 pairs contribute to 9 handshakes/person.
Please elaborate.

Thanks :)
Here's another way to look at it.

There are ten people in the room shaking hands, but each person does not shake hands with himself or herself. The person shakes hands with the other nine people in the room.

So from each of the ten people's perspectives, that person shakes hands with nine other people.

So in a way it seems as if this is what went on. 10 people x 9 handshakes = 90 handshakes total

But wait. When one person shakes hands, another person has also shaken hands. So, one handshake takes care of one handshake for each of two people. So there are not really 90 handshakes. There are only 45.
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by nikhilgmat31 » Mon Aug 31, 2015 5:17 am
[email protected] wrote:Hi ahmedbari.ace,

Here is how you can 'visualize' 10 people each shaking hands with one another.

Let's name the people A,B,C,D,E F,G,H,I,J

The 45 handshakes would be....

AB,AC,AD,AE,AF,AG,AH,AI,AJ
BC,BD,BE,BF,BG,BH,BI,BJ
CD,CE,CF,CG,CH,CI,CJ
DE,DF,DG,DH,DI,DJ
EF,EG,EH,EI,EJ
FG,FH,FI,FJ
GH,GI,GJ
HI,HJ
IJ

You'll notice that there are 9+8+7+6+5+4+3+2+1 = 45 handshakes and that each letter appears in exactly 9 different handshakes.

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Hi Rich, I have a little confusion here. One way is using Mitch's way to divide 10C2 by 3 & reach to answer of 15.

However I am trying it in different way little similar to yours.

possible combinations of HandShakes are :-

A - BCD - 3 new Handshakes
B - CD - 2 new Handshakes
C - D - 1 new Handshakes
D - he is already done - 0 new Handshakes

E - FGH - 3 new Handshakes
F - GH - 2 new Handshakes
G - H - 1 new Handshakes
H - he is already done - 0 new Handshakes

I - J - 1 new Handshake - Now problem here is that I has no other person for handshake since all else J has done 3 handshakes

J - I - 1 new Handshake - Same is the problem with J


so total handshakes comes out to be = 3 + 2 + 1 + 0 + 3 + 2 + 1 + 0 + 1 +1

= 14

Please suggest where I missed or can we assume at I & J handshake for 3 times which doesn't makes a practical sense.
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by GMATGuruNY » Mon Aug 31, 2015 7:03 am
[email protected] wrote:
GMATGuruNY wrote:
The total number of pairs that can be formed from 10 people = 10C2 = 45.
If all of these pairs shake hands, each person in the room will shake hands with EVERY OTHER PERSON in the room.
The result will be 9 HANDSHAKES PER PERSON.
Hi Mick

How do 45 pairs contribute to 9 handshakes/person.
Please elaborate.

Thanks :)
Let the 10 people be A, B, C, D, E, F, G, H, I, and J.
From these 10 people, a total of 45 pairs can be formed, as shown in my post above.

Person A is included in the following pairs:
AB, AC, AD, AE, AF, AG, AH, AI and AJ.
Thus, if all 45 pairs shake hands, A will shake hands with 9 other people.

This reasoning will hold true for B, C, D, E, F, G, H, I and J.
If all 45 pairs shake hands, B will shake hands with 9 other people.
If all 45 pairs shake hands, C will shake hands with 9 other people.
And so on.

Implication:
If all 45 pairs shake hands, each person will shake hands with 9 other people.
Dividing by 3, we get:
If 15 pairs shake hands, each person will shake hands with 3 other people.
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by Max@Math Revolution » Wed Sep 02, 2015 9:33 pm
Forget conventional ways of solving math questions. In PS, IVY approach is the easiest and quickest way to find the answer.


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

A) 15
B) 30
C) 45
D) 60
E) 120

Since handshakes are made betwen two people, and we have 10 people overall, 10C2 = 45. However this is a case where 1 person makes handshake with the other 9, while the question asks for the number of handhsakes 1 person makes with other 3. Therefore, 45:9=x:3, x=15. The answer is A.


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by nikhilgmat31 » Wed Sep 02, 2015 11:37 pm
GMATGuruNY wrote:
[email protected] wrote:
GMATGuruNY wrote:
The total number of pairs that can be formed from 10 people = 10C2 = 45.
If all of these pairs shake hands, each person in the room will shake hands with EVERY OTHER PERSON in the room.
The result will be 9 HANDSHAKES PER PERSON.
Hi Mick

How do 45 pairs contribute to 9 handshakes/person.
Please elaborate.

Thanks :)
Let the 10 people be A, B, C, D, E, F, G, H, I, and J.
From these 10 people, a total of 45 pairs can be formed, as shown in my post above.

Person A is included in the following pairs:
AB, AC, AD, AE, AF, AG, AH, AI and AJ.
Thus, if all 45 pairs shake hands, A will shake hands with 9 other people.

This reasoning will hold true for B, C, D, E, F, G, H, I and J.
If all 45 pairs shake hands, B will shake hands with 9 other people.
If all 45 pairs shake hands, C will shake hands with 9 other people.
And so on.

Implication:
If all 45 pairs shake hands, each person will shake hands with 9 other people.
Dividing by 3, we get:
If 15 pairs shake hands, each person will shake hands with 3 other people.
Hi Mitch,

we are able to reach at 10C2 = 45 if there is no limitation of exactly 3 handshakes per person.

After that, we know 1 person shakes hand with 3 other person. My concern is that the last person I & J don't have enough persons to shake hand 3 times since all others have already did it 3 times.

Please elaborate with 15 combinations of handshakes
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by [email protected] » Thu Sep 03, 2015 9:10 am
Hi nikhilgmat31,

From what you describe, it sounds like you're thinking in terms of 'groups of 4', meaning if you had 4 people....

A,B,C,D

Each person in this group would shake hands with each of the 3 other people in the group and you'd have the following 6 handshakes...

AB
AC
AD
BC
BD
CD

The prompt does NOT state this restriction though; it states that there are 10 total people and that each person shakes hands with 3 others (in this scenario, the handshakes are 'random', and not restricted to a 'group of 4').

However, using this idea, and the answer choices (which are relatively 'spread out'), you could quickly deduce the correct answer... If a group of 4 people would generate 6 handshakes, then a different group of 4 people would generate another 6 handshakes, giving us a total of 12. The 2 extra people that we haven't accounted for yet are NOT enough to generate another 6 handshakes, so the total number of handshakes MUST be LESS than 18... and there's only one answer that fits.

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by nikhilgmat31 » Thu Sep 03, 2015 8:22 pm
Hi Rich,

I am trying it in different way little similar to yours.

possible combinations of HandShakes are :-

I didn't make combinations of 4 but I am picking 1 guy & note down his handshakes as below

A - BCD - 3 new Handshakes
B - CD - 2 new Handshakes as he has already done with A
C - D - 1 new Handshakes as he has already done with A and B
D - he is already done - 0 new Handshakes as he has already done with A,B and C

E - FGH - 3 new Handshakes
F - GH - 2 new Handshakes as he has already done with E
G - H - 1 new Handshakes as he has already done with E and F
H - he is already done - 0 new Handshakes as he has already done with E,F,G

I - J - 1 new Handshake - Now problem here is that I has no other person for handshake since all else J has done 3 handshakes

J - I - 0 new Handshake - J is already done with I


so total handshakes comes out to be = 3 + 2 + 1 + 0 + 3 + 2 + 1 + 0 + 1 + 0


= 13

Please suggest where I missed or can we assume at I & J handshake for 3 times which doesn't makes a practical sense.
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