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500 ps test6 #20

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500 ps test6 #20

by dunkin77 » Thu Apr 05, 2007 3:12 pm
Hi,

I though the answer is 10P9 C) but the correct answer is D).....

. If 10 persons meet at a reunion and each person shakes hands exactly once with each of the others, what is the total number of handshakes?
(A) 10•9•8•7•6•5•4•3•2•1
(B) 10•10
(C) 10•9
(D) 45
(E) 36
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by vk.neni » Thu Apr 05, 2007 7:47 pm
Hi, This is the sum of a progression kind of problem.

The 1st person would shake with 9 others, the 2nd would do with 8 others etc. So, the final answer would 9+8+...+1 = 9*(9+1)/2 = 45.
Answer D.
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by jayhawk2001 » Thu Apr 05, 2007 8:56 pm
Simply put, we are asked to find the number of ways 10 people can
shake hands with each other.

10C2 is the answer
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by dunkin77 » Thu Apr 05, 2007 9:28 pm
Thank you.

I am always confursed using P and C... (if you could a bit explain, that would be soo great...)

In this case, 10C2, 10 is obviously 10 people, but where the number 2 come from? meaning avoiding duplication?

Thanks!
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by jayhawk2001 » Thu Apr 05, 2007 9:44 pm
dunkin77 wrote:Thank you.

I am always confursed using P and C... (if you could a bit explain, that would be soo great...)

In this case, 10C2, 10 is obviously 10 people, but where the number 2 come from? meaning avoiding duplication?

Thanks!
You might want to check out the GMAT resource wiki. It has a lot of
useful info in there that I have personally found helpful

https://www.beatthegmat.com/wiki/doku.ph ... _resources
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by Cybermusings » Sat Apr 07, 2007 12:21 am
If 10 persons meet at a reunion and each person shakes hands exactly once with each of the others, what is the total number of handshakes?
(A) 10•9•8•7•6•5•4•3•2•1
(B) 10•10
(C) 10•9
(D) 45
(E) 36

For a handshake 2 people are required. Hence 2 people can be selected from a group of 10 in 10C2 ways

= 45

Hence Choice D
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by BTGmoderatorRO » Sun Oct 29, 2017 9:45 am
since each person shakes hands exactly once with each of the others, this is just the same as grouping two people together without considering the order.
e.g in {a,b,c,d}, the number of handshakes will be 6. i.e {a,b], {a,c}, {a,d}, {b,c}, {b,d} and {c,d}.
This is given by $$n_{Cr}$$ , where n is the total number of people and r is the number of people shaking hands once.
Total number of handshakes is thus given as
$$_{10_{C2}=\ \frac{10!}{\left(10-2\right)!2!}\ =\ \frac{\left(10\cdot9\cdot8!\right)}{8!\cdot2!}=\frac{\left(10\cdot9\right)}{2}=45}$$
$$Therefore,\ there\ will\ be\ 45\ handshakes\ if\ two\ people\ do\ not\ shake\ hands\ with\ each\ other\ more\ than\ one\ time$$
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by Brent@GMATPrepNow » Mon Oct 30, 2017 8:47 am
If 10 persons meet at a reunion and each person shakes hands exactly once with each of the others, what is the total number of handshakes?

(A) 10!
(B) 10*10
(C) 10*9
(D) 45
(E) 36
Another approach:

Once everyone has shaken hands, ask each of the 10 people, "How many people did you shake hands with?"
We'll find that EACH PERSON shook hands with 9 people, which gives us a total of 90 handshakes (since 10 x 9 = 90).

From here 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 90 by 2 to get 45

Answer: D

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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