This is a classic combination problem, albeit with large numbers.

If the order mattered (a permutation), the answer would be:

12 x 11 x 10 x 9 = 11,880

However, since order doesn't matter, you need to divide by n!, where "n" is the number of items in your numerator (in this case, the numerator is 12x 11 x 10 x 9, so that's 4 items).

So, divide 11,880 by 4! (4! = 24) to get 495.

And yes, I know that I modified the combination formula a bit, but on virtually every problem, I think my adapted formula is a little easier than the "official" formula.

Hope this helps!

To solve this problem let's begin by reviewing the basics of factorials, which are shown with an exclamation point. When we see this, we must basically "multiply by the countdown." By way of example:

6! = (6 x 5 x 4 x 3 x 2 x 1), and 4! = (4 x 3 x 2 x 1)

We should also recognize how factorials work in division. For example,

6!/4! can be simplified, because the two values have many terms in common which cancel out:

(6 x 5 x 4 x 3 x 2 x 1)/(4 x 3 x 2 x 1) = (6 x 5)/1 = 30



Now, for the problem at hand, we must apply the subset combination formula: Number of subset combinations = X!/Y!(X-Y)!

Where X = total number of items (in this case, 12)
and Y = number of items chosen from the set (in this case, 4):

Number of subset combinations = X!/Y!(X-Y)!

Number of subset combinations = 12!/4!(12-4)! = 12!/(4!)(8!) = 12!/(8!)(4!)

Many terms cancel out between the 12! in the numerator and the 8! in the denominator, leading to the following, simplified equation:

Number of subset combinations = (12 x 11 x 10 x 9)/(4 x 3 x 2 x 1)

To further simplify, we should notice that the 12 in the numerator cancels out the 4 x 3 in the denominator, leaving us with this:

Number of subset combinations = (11 x 10 x 9)/(2 x 1) = 990/2 = 495


I hope that's helpful! Let me know--thanks!
[/u]

vertigo05 wrote: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 restaurant, in how many different ways can company select the 4 sites needed for the new restaurant if the order of selection doesn't matter.
A. 48
B. 288
C. 495
D. 990
E. 11880

Whenever we're choosing subgroups out of a big group, we want to think about either combinations or permutations.

Next, we want to ask ourselves if order matters. Here it clearly doesn't ("if the order of selection doesn't matter"), which means that we have a combinations question.

In general:

nCk = n!/k!(n-k)!

in which:

n = total number of objects available
k = number of objects included in the subgroup

So, applying the forumla to this question:

12C4 = 12!/4!(12-4)! = 12!/4!8! = 12*11*10*9/4*3*2*1

(shortcut applied above: cancel out the bigger factorial on the bottom with part of the factorial in the top; since 12! = 12*11*10*9*8*7*6*5*4*3*2*1, we can rewrite it as 12*11*10*9*8! and then cancel out the 8! on top and bottom of the fraction)

= 12*11*10*9/4*3*2 = 11*10*9/2 = 99*5 = 495... choose (C).

Hah! We were all typing at the same time.

I guess we can't resist math problems with no replies!

Thnx to all..nice explanations.

Is there something wrong? Order must be a matter.