GMAT Prep- A fast food company

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vertigo05: Senior | Next Rank: 100 Posts; Posts: 62; Joined: Fri May 30, 2008 5:37 am

by vertigo05 » Tue May 05, 2009 10:18 am

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

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Source: Beat The GMAT — Problem Solving | Tue May 05, 2009 10:18 am

VP_Jim: GMAT Instructor; Posts: 1223; Joined: Thu May 01, 2008 3:29 pmFollowed by: 15 members; Location: Los Angeles, CA; Thanked: 185 times

by VP_Jim » Tue May 05, 2009 11:28 am

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!

Jim S. | GMAT Instructor | Veritas Prep

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stevestein: GMAT Instructor; Posts: 5; Joined: Mon Mar 30, 2009 11:30 am; Thanked: 3 times; GMAT Score:770

by stevestein » Tue May 05, 2009 11:28 am

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]

Steve Stein
PowerScore GMAT Instructor

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Stuart@KaplanGMAT: GMAT Instructor; Posts: 3225; Joined: Tue Jan 08, 2008 2:40 pmFollowed by: 614 members; Location: Toronto; Thanked: 1710 times; GMAT Score:800

Re: GMAT Prep- A fast food company

by Stuart@KaplanGMAT » Tue May 05, 2009 11:33 am

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).

Stuart Kovinsky | Kaplan GMAT Faculty | Toronto

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Stuart@KaplanGMAT: GMAT Instructor; Posts: 3225; Joined: Tue Jan 08, 2008 2:40 pmFollowed by: 614 members; Location: Toronto; Thanked: 1710 times; GMAT Score:800

by Stuart@KaplanGMAT » Tue May 05, 2009 11:34 am

Hah! We were all typing at the same time.

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VP_Jim: GMAT Instructor; Posts: 1223; Joined: Thu May 01, 2008 3:29 pmFollowed by: 15 members; Location: Los Angeles, CA; Thanked: 185 times

by VP_Jim » Tue May 05, 2009 12:52 pm

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

Jim S. | GMAT Instructor | Veritas Prep

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vertigo05: Senior | Next Rank: 100 Posts; Posts: 62; Joined: Fri May 30, 2008 5:37 am

by vertigo05 » Wed May 06, 2009 8:42 am

Thnx to all..nice explanations.

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tanviet: Legendary Member; Posts: 1404; Joined: Tue May 20, 2008 6:55 pmFollowed by: 2 members; Thanked: 18 times

by tanviet » Sun Apr 29, 2012 6:28 am

Is there something wrong? Order must be a matter.

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