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105 hand-shakes

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by sanju09 » Wed Feb 11, 2009 5:02 am
Persons gathered in a meeting shook hands once among each other, with no one to spare. An un-invited was happy counting the total number of hand-shakes that took place that way, and finally he murmered: "Wow! 105 hand-shakes!" How many persons were attending the meeting there?
A. 105
B. 35
C. 25
D. 15
E. 12
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by mals24 » Wed Feb 11, 2009 5:30 am
I guess D

Let the total number of guests = N

Total number of handshakes = N(N-1)/2 = 105

N(N-1) = 210

N = 15
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by sureshbala » Wed Feb 11, 2009 5:40 am
If there are n persons in the party, since each person shakes hand with all the others exactly once, the total number of shake hands will be nC2.

Given nC2 = 105. Hence n = 15
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by ellexay » Wed Feb 11, 2009 5:41 pm
Suresh,

How did you find out the value of N in NC2=105? Is there a formula?

Thanks!
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by DeepakR » Wed Feb 11, 2009 5:53 pm
Ncr=n!/(n-r)!*r! so Nc2=n!/(n-2)!*2!=n*(n-1)*(n-2)!/(n-2)!*2!
hence n*(n-1)/2=210 solving we get n=15

- Deepak
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by ven4gmat » Thu Feb 12, 2009 1:48 am
If n is the number people in the party, then the total number of shakehands will be nC2.

given nC2 = 105

i.e n(n-1)/2 = 105.

So n =15
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