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In a certain conference, if all the n attendees shake hands 105 times,

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Math Revolution GMAT Instructor
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In a certain conference, if all the n attendees shake hands 105 times, [#permalink]

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New post 18 Aug 2017, 01:56
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A
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C
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E

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Question Stats:

83% (00:47) correct 17% (01:26) wrong based on 65 sessions

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In a certain conference, if all the n attendees shake hands 105 times, what is the value of n?

A. 10
B. 12
C. 15
D. 16
E. 18

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Re: In a certain conference, if all the n attendees shake hands 105 times, [#permalink]

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New post 18 Aug 2017, 02:01
1
MathRevolution wrote:
In a certain conference, if all the n attendees shake hands 105 times, what is the value of n?

A. 10
B. 12
C. 15
D. 16
E. 18


n(n-1)/2 = 105
n(n-1) = 210
(n-15)(n+14) = 210
n = 15

Hence option C is correct
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Re: In a certain conference, if all the n attendees shake hands 105 times, [#permalink]

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New post 18 Aug 2017, 04:39
I solved this question in 1,5 minutes, but with plug in method.

Since this one is a combination problem, so we should determine what nCr will have 105 in result.

MathRevolution, do we have more straightforward approach?
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Re: In a certain conference, if all the n attendees shake hands 105 times, [#permalink]

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New post 18 Aug 2017, 05:02
MathRevolution wrote:
In a certain conference, if all the n attendees shake hands 105 times, what is the value of n?

A. 10
B. 12
C. 15
D. 16
E. 18


Hi..

Ofcourse the Q should mention that each shake hands with others once.

So choosing 2 out of n is nC2=\(\frac{n!}{(n-2)!2!}=\frac{n*(n-1)*(n-2)!}{(n-2)!*2!}=\frac{n*(n-1)}{2}=105....\)...
\(n(n-1)=210=2*3*5*7=3*5*2*7=15*14\)
So n is 15
C
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Re: In a certain conference, if all the n attendees shake hands 105 times, [#permalink]

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New post 20 Aug 2017, 19:14
nC2 = 105.
n(n-1)/(1*2) = 105
n(n-1) = 210
n(n-1) = 15*14

Thus n = 15.

Ans: C
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Re: In a certain conference, if all the n attendees shake hands 105 times,   [#permalink] 20 Aug 2017, 19:14
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