PS

[lionheart]

A fast food company plans to build 4 new restaurants. If there are 12 sites that satisfy the company's criteria for location of new restaurants, in how many different ways can the company select the 4 sites needed for the new restaurants if the ordeer of selection does not matter?

A) 48
B) 288
C) 495
D) 990
E) 11880

[GMAT700]

12!/4!8! (MGMAT's way of doing this, you should take the prep course =p. Judging by the fact that you don't know how to do this problem, I don't think you are or were a past student =p)

12*11*10*9*8...stop divided by 4*3*2*1, the 8! cancels out w/ the 8! in the numerator.'

cancel out the 12 in the numerator and the 4*3 in the denominator.

you're left with 11*10*9 / 2

990 / 2 = 495 different groups of 4 can be selected from 12.

[RonPurewal]

yeah, this is a straight combination problem, perfectly suited to our 'anagram method' (the method that produces the formula 12!/(4!8!) referenced in the previous post).

if you don't know that method, or prefer the fundamental counting principle instead, you can also do the following:
12 (choices for first location) x 11 (choices for second location) x 10 (choices for third location) x 9 (choices for fourth location)
but then because the order doesn't matter, you have to divide out by the 'redundancy factor', which is 4 x 3 x 2 x 1 (there are this many ways to arrange the same four locations, so 12 x 11 x 10 x 9 will give this many repetitions of each set of four locations)

so the correct answer, then, is
(12 x 11 x 10 x 9) / (4 x 3 x 2 x 1), as arrived at in the other method as well.