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Combinations Question

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by Nole23 » Sun Feb 22, 2009 10:51 am
Is there a shortcut for solving the combinations question: 9! / 5!(4!) as opposed to actually multiplying everything out and then dividing?

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by Bidisha_800 » Sun Feb 22, 2009 1:38 pm
=9*8*7*6*5!/5!*4!

=9*8*7*6/4*3*2

=18*7

=126
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by Nole23 » Sun Feb 22, 2009 3:50 pm
Could you elaborate on your math a little bit?
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by Brent@GMATPrepNow » Sun Feb 22, 2009 4:16 pm
To evaluate mCr, write the first r values of m! in the numerator and r! in the deniminator.

Eg: 6C2 = (6x5)/2!
Eg: 8C3=(8x7x6)/3!
Eg: 10C4 = (10x9x8x7)/4!

To evaluate these, do not calculate the products in the numerator and denominator and then simply. Rather, simplify before multiplying.

Eg: 9C3 = (9x8x7)/3! = (9x8x7)/(3x2x1)
We have 9 in the numerator and 3 in the denominator. Simply to get: (3x8x7)/(1x2x1)
Similarly, we have 8 in the numerator and 2 in the denominator. Simply to get: (3x4x7)/(1x1x1) = 84

Hope that helps.
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