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Police cars probability

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by gmattesttaker2 » Tue Dec 31, 2013 7:08 pm
Hello,

Can you please assist with this:

The Springville Police Department has C police cars available. When one car breaks down, 28 different six-car combinations are available to patrol the streets of Springville. What is the value of C?

Answer Choices:

A) 11
B) 10
C) 9
D) 8
E) 7

OA: C


My approach was as follows:

(C-1)(C-2)(C-3)(C-4)(C-5)(C-6) / 6! = 28

I tried to solve this by looking at the answer choices and taking values for C till I got the correct value on the left hand side. I was wondering if there is a better way to solve this problem.

Thanks a lot,
Sri
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by GMATGuruNY » Wed Jan 01, 2014 6:34 am
gmattesttaker2 wrote:Hello,

Can you please assist with this:

The Springville Police Department has C police cars available. When one car breaks down, 28 different six-car combinations are available to patrol the streets of Springville. What is the value of C?

Answer Choices:

A) 11
B) 10
C) 9
D) 8
E) 7

OA: C
We can PLUG IN THE ANSWERS, which represent the total number of cars.

Answer choice C: 9
After 1 car breaks down, the number of cars available for patrol = 9-1 = 8.
From these 8 cars, the number of combinations of 6 that can be formed = 8C6 = (8*7*6*5*4*3)/(6*5*4*3*2*1) = 28.
Success!

The correct answer is C.
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by Brent@GMATPrepNow » Wed Jan 01, 2014 7:56 am
gmattesttaker2 wrote: The Springville Police Department has C police cars available. When one car breaks down, 28 different six-car combinations are available to patrol the streets of Springville. What is the value of C?

Answer Choices:

A) 11
B) 10
C) 9
D) 8
E) 7
I'd also use Mitch's approach of testing answers. I just want to add a few things.

First, if anyone is interested, we have a free video on calculating combinations in your head: https://www.gmatprepnow.com/module/gmat-counting?id=789

Second, when calculating combinations (using the fast method described in our video), it's sometimes useful to apply the following rule: nCr = nC(n-r). In other words, "n choose r" is equal to "n choose n - r"
So, for example: 10C7 = 10C3
12C9 = 12C3
8C7 = 8C1
etc.

So, when we test whether or not 8C6 = 28, we can test 8C2 instead.
8C2 = (8)(7)/(2)(1) = 28
Since 8C2 = 28, we know that there were 9 cars before one car broke down.

Cheers,
Brent
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