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Probability of cars

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Probability of cars

by gmatNooB8787 » Wed May 02, 2012 6:49 pm
I am not able to under stand the solution of the following problem:

The Carson family will purchase three used cars. There are two models of cars available, Model A and Model B, each of which is available in four colors: blue, black, red, and green. How many different combinations of three cars can the Carsons select if all the cars are to be different colors?
options -> 24,32,48,60,192.

According to the answer : we first check how each slot for a car can be filled in _ _ _ => 8 * 6 * 4;
hence the answer should be :192

However, the solution further says ,
The order in which the purchases are made is not important so we must divide by the factorial of the number of choices to eliminate over-counting:
( 8 * 6 * 4 ) / 3! = 32, which is the answer.


I am not able to understand this last part. In calculating 192 we already take care of order , then why divide by 3! again ?
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by aneesh.kg » Wed May 02, 2012 7:12 pm
I will list three methods of solving this problem in my order of preference.

Method 1:
Select 3 colours out of 4 colours (4C3) and then select one model-type from each of those colours (2C1 three times).
Total number of ways: 4C3 * 2C1 * 2C1 * 2C1 = 32

Method 2:
Select any three cars from the 8 cars (8C3) and subtract those ways in which there is a pair of cars with same colour. How many ways of selection are there in which two cars of the same colour are chosen? First select a pair of such cars (4C1) and then select one car from the remaining six cars (6C1).
Total number of ways: 8C3 - (4C1)*(6C1) = 32

Method 3:
Select any one of the 8 cars (8C1) and then select any one of the remaining 6 cars which are not of the previous colour (6C1) and then select another car which is not of the previous two colours (4C1). Then, since we have arranged instead of selecting, we must divide the total number of ways to nullify the arrangement.
Let's take an example:
In this way of selection, you will end up with
Blue-A Black-B Green-A
Green-A Blue-A Black-B
Green-A Black-B Blue-A
...and so on.
If you notice, all these comprise the same set of cars. We are ending up arranging these three cars. Do we need to do that? We don't. All we need is a set of three cars. In each correct way of selection, how many arrangements are happening? 3!.
In every one correct way of making a selection, we are selecting 3! number of teams unnecessarily.
So if 'x' is the correct ways of arrangement,
8C1 * 6C1 * 4C1 = 3! times 'the correct ways of arrangement'
8 * 6 * 4 = 3! * x
x = (8 * 6 * 4)/3! = 32

We follow the same principle in questions such as:
How many arrangements of the word MATHEMATICS are possible?

A similar concept:
https://www.beatthegmat.com/important-pe ... tml#469282
Last edited by aneesh.kg on Wed May 02, 2012 9:12 pm, edited 2 times in total.
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by spartacus1412 » Wed May 02, 2012 7:47 pm
The question places a restriction that no two cars of the smae colour are to be selected. It puts no restriction on the model of the car.
Hence, the colour of the 3 cars from 4 available colurs can be selected by 4C3 ways.
but each car can be either model A or Model B.

2*2*2* 4C3
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by Stuart@KaplanGMAT » Wed May 02, 2012 11:25 pm
gmatNooB8787 wrote:
The order in which the purchases are made is not important so we must divide by the factorial of the number of choices to eliminate over-counting:
( 8 * 6 * 4 ) / 3! = 32, which is the answer.


I am not able to understand this last part. In calculating 192 we already take care of order , then why divide by 3! again ?
Hi!

Lots of good solutions already posted, so I'm just going to focus on your actual question.

When we made the calculation to get 192, we were working on the basis that order DID matter. If you think about those possibilities, you'll see that we've over counted.

For example, if the first car was a red "A", then we've eliminated red and the next car could be blue "A". Now we've eliminated blue, so the last car could be yellow "A".

However, another group that we've counted is the first car being blue "A", the second car yellow "A" and the third car red "A".

Using the above method, we've counted these two strings as two different car choices. But, since they're both "red A, blue A, yellow A", they're actually the same choice.

Since there are 3 items on the list, we have to divide by 3! to get rid of all of the duplicate strings.

Just in case that's not clear enough, let's look at a simpler example:
There are 4 cars available for purchase: one is green, one blue, one red and one yellow. How many different groups of 3 cars can be chosen?
Using the method you propose, the answer would be:

4 for the first car, 3 for the second car, 2 for the 3rd car, so 4*3*2 = 24

However, we've counted:

GBR
GRB
BRG
BGR
RGB
RBG

as 6 different options, when in fact those are all identical. So, the correct answer would be:

4*3*2/3! = 24/6 = 4

In fact, the difference between adding that 3! in the divisor is the difference between the formulas for permutations and combinations!

If order DOES matter, use permutations:

nPk = n!/(n-k)!

If order DOES NOT matter, use combinations:

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