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Reverse Combinatorics Problem - GMAT Prep

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by Rahul@gurome » Wed Oct 27, 2010 7:31 pm
Solution:
Let the total number of members be x.
So xC2 = 190.
Or x(x-1)/2 = 190.
Or x(x-1) = 380.
Solving you get x = 20.
The correct answer is A.
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by tlt2372 » Wed Oct 27, 2010 8:16 pm
Rahul@gurome wrote:Solution:
Let the total number of members be x.
So xC2 = 190.
Or x(x-1)/2 = 190.
Or x(x-1) = 380.
Solving you get x = 20.
The correct answer is A.
How did you move from xC2 to x(x-1)/2?

(I am really terrible at Combinatorics) :(
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by Rahul@gurome » Wed Oct 27, 2010 8:30 pm
xC2 is x!/{(x-2)!*2!}.
Now x! is x(x-1)(x-2)....2*1 = x(x-1)(x-2)!.
So x!/{(x-2)!*2!} is {x(x-1)(x-2)!}/{(x-2)!*2!} = x(x-1)/2! = x(x-1)/2.
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by tlt2372 » Wed Oct 27, 2010 8:47 pm
Rahul@gurome wrote:xC2 is x!/{(x-2)!*2!}.
Now x! is x(x-1)(x-2)....2*1 = x(x-1)(x-2)!.
So x!/{(x-2)!*2!} is {x(x-1)(x-2)!}/{(x-2)!*2!} = x(x-1)/2! = x(x-1)/2.
Gotcha! Thanks!
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