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Hard PS - looking for faster way to solve

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Manager
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Hard PS - looking for faster way to solve [#permalink] New post 01 Nov 2006, 17:10
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Question Stats:

100% (01:07) correct 0% (00:00) wrong based on 0 sessions
Took me 6 mins to solve this question manually, is there a faster way to solve it mathematically ?

In a class comprising boys and girls, there were 45 hand shakes amongst the girls and 105 hand shakes amongst the boys. How many hand shakes took place between a boy and a girl, if each member of the class shook hands exactly once with every other student in the class?
25
15
150 (Correct Answer)
300
24

TIA.
Director
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 [#permalink] New post 01 Nov 2006, 17:34
if g is no of girls: g(g-1)/2 = 45 --> g=10
if b is no of boys: b(b-1)/2 = 210 --> b=15

handshakes between b and g-->15*10 = 150
Director
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 [#permalink] New post 01 Nov 2006, 18:18
Juaz,

Where did you come up with the equations?
Manager
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 [#permalink] New post 02 Nov 2006, 06:04
Let G = # girls.

Then the number of handshakes among girls is G choose 2 (I write GC2). So GC2 = 45. A useful thing to remember is that XC2 = X*(X-1)/2, so G = 10. Similarly, B = # boys so BC2 = 105 or B = 15. The # handshakes is then B*G = 150.
(Edit - I just saw that this is juaz' solution as well)
VP
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 [#permalink] New post 02 Nov 2006, 20:36
45 = 10C2 girls

105 = 15C2 boys

So 10 girls and 15 boys

Total handshakes 10*15 =150

**************************************

it takes 2 people for 1 handshake

45 = nC2

n*(n-1) = 90
n=10

||ly the # of boys = 15
Director
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 [#permalink] New post 05 Nov 2006, 19:08
Will/can someone explain where this equation comes from:
XC2 = X*(X-1)/2
Other than this, when is it used?

Sorry to bring this up again.
  [#permalink] 05 Nov 2006, 19:08
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