Permutations seat arrangements

[brendan]

Homework Statement

9 people, 2 cars can hold 5 people each. only 3 people have licenses.

Homework Equations

How many different ways can the people be seated into cars?

The Attempt at a Solution

There are nine people. So number of seats is not a problem.

We must find out how many groups of three people with licenses there can be.

nPr = n!(n-r)!
n = 9
r = 3
nPr= 9!/(9-3)! = 504


How's that look?

 

[HallsofIvy]

Homework Statement

9 people, 2 cars can hold 5 people each. only 3 people have licenses.

Homework Equations

How many different ways can the people be seated into cars?

The Attempt at a Solution

There are nine people. So number of seats is not a problem.

We must find out how many groups of three people with licenses there can be.

How did you arrive at that? The problem is to divide the 9 people into 2 groups. And, since there are only two cars, in each possibility one of the people with a license is irrelevant.

(n-r)!
n = 9
r = 3
nPr= 9!/(9-3)! = 504
Choose one of the three people with licenses to drive the first car. There are 3! ways to do that. Choose one of the two remaining people with licenses to drive the the second car. There are now 7 people to be distributed between the two cars. That can be done by putting 4 people in car A and 3 in car B or 3 people in car A and 4 people in car B. If you were to put 4 people in car A you can think of that as permutations of 4 As and 3Bs. If you put 4 people in car B that is a permutation of 3As and 4Bs but is the same number. Add those two numbers.
How's that look?

 

[brendan]

So It works out to be

nPr = n!(n-r)!
n = 4
r = 3
nPr= 4!/(1)! = 24 +


nPr = n!(n-r)!
n = 4
r = 3
nPr= 4!/(1)! = 24


= 48?

 

[brendan]

Just been looking at this problem again.

If there are 3 licensed drivers isn't there 3! (six times) they can be arranged ?

Which would leave 6 pasengers arranged into two cars say

3 in car A (+ 2 drivers) and 3 in car B (+ 1 driver) = nine all up

Would you than calculate:

3!(6C3) + 3!(6C3)

nCr = n!(n-r)!r!
> n = 6
> r = 3
> nCr= 6!/(3)!3!
>
> =20 * 3!
=120
>
+
>
> nCr = n!(n-r)!r!
> n = 6
> r = 3
> nCr= 6!/(3)!3!
= 20 * 3!
= 120


= 240 ?

This is very confusing AAARRRHHH !

Thanks,
Brendan

 

[HallsofIvy]

Just been looking at this problem again.

If there are 3 licensed drivers isn't there 3! (six times) they can be arranged ?

If you say "and then take the first two to be drivers of the two cars (in order)" then, yes, that is correct.

Which would leave 6 pasengers arranged into two cars say

Be careful. There are not three cars so one of the people with a license has to be included amoung the 7 passengers.

3 in car A (+ 2 drivers) and 3 in car B (+ 1 driver) = nine all up

You are assuming that the licensed person who is not driving will be in car A. That is not given.

Would you than calculate:

3!(6C3) + 3!(6C3)

nCr = n!(n-r)!r!
> n = 6
> r = 3
> nCr= 6!/(3)!3!
>
> =20 * 3!
=120
>
+
>
> nCr = n!(n-r)!r!
> n = 6
> r = 3
> nCr= 6!/(3)!3!
= 20 * 3!
= 120


= 240 ?

This is very confusing AAARRRHHH !

Thanks,
Brendan

 

[brendan]

I see what you mean.

There are 3! drivers = 6
But after the 2 drivers are selected there are still 7 people to be allocated.

Therefore there are 7C4 = 35 combinations of people.

So we multiply the number of passenger combinations by the driver combinations to get

35 X 6 = 210.

Personally I would tell them all just to get in however they wanted!
regards
Brendan