Question 936069: Please help me solve this:
Jodi is parking seven different types of vehicles side by side facing a display window at the dealership where she works, how many ways can she park them so that the convertible is not next to the subcompact?
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! it appears the formula will be:
8! - 7! * 2!
you have 8 different type vehicles.
you don't want 2 of the types to be together.
the simplest way to calculate this is:
find the total number of arrangements without any restrictions.
find the total number of arrangements assuming the two different types of vehicles have to be together.
subtract the number of ways the 2 types of vehicles have to be together from the total number of ways they can be together without any restrictions and you have the total number of ways they can be together when the two types of vehicles are not next to each other.
the total number of ways you can arrange them without any restrictions is 8!.
the total number of ways you can arrange them assuming the two different types of vehicles have to be together is 7! * 2!.
what is happening here is that you are treating the 2 as one which results in 7! rather than 8!. you also need to determine that, within their pairing, they have 2! different types of ways they can be shown, i.e. ab and ba. that's why you multiply the 7! by 2!.
to find the number of ways you can arrange them assuming the two different types of vehicles cannot be together becomes 8! - 7! * 2!.
i'll show you with a simpler example how this works so you can feel more comfortable with the solution.
let's make it very simple.
assume you have 3 different type vehicles.
assume the types are a, b, and c.
the number of different ways you can arrange them without any restrictions would be 3! = 6.
the 6 different arrangements would be:
abc
acb
bac
bca
cab
cba
now assume that types a and b have to be together.
the number of different ways you can arrange them where a and b have to be together would be based on the formula of:
2! * 2! = 4
your 3! became 2! because you treated treated the 2 different types that have to be together as one, and you multiplied that by 2! because those 2 vehicles could be arranged 2! ways when they are together.
let's take a look at our unrestricted number of arrangements and mark the ones where a and b have to be together.
the total should be 4.
the 6 unrestricted arrangements are shown below.
the arrangements that have a and b together are marked.
abc ****
acb
bac ****
bca
cab ****
cba ****
the arrangements where a and b are not together are not marked.
there are 2 of them.
they are the difference between the ones that are marked and the total number of arrangements without any restrictions.
6 - 4 = 2
there are 2 arrangements where a and b are not together.
the same concept is used with the larger numbers.
the total number of unrestricted arrangements for 8 different type vehicles is 8!.
the total number of arrangements where 2 types of those vehicles have to be together is 7! * 2!.
we treat the 2 types that have to be together as 1 which results in 7! rather than 8!. we also take into account that, even they are together, they can be arranged in different ways while they are together. since there are 2 of them that have to be together, we know that they can be arranged in 2! ways which is why you have 7! * 2!.
the difference is the number of arrangements where the two types of vehicles will not be together.
the solution is 8! - 7! * 2! which is equal to 30240.
there are 30240 different arrangements where the two types of vehicles will not be together.
i'm pretty sure this is correct.
these types of problems are difficult to confirm they are correct because there are so many different possibilities.
use of a simple example is a good way to confirm, which is what i did.
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