The problem involves arranging 9 people into 2 cars that can each hold 5 people, with the constraint that only 3 of the individuals have driving licenses. Participants are exploring the various ways to seat the individuals based on these conditions.
The discussion is ongoing, with multiple interpretations of how to approach the problem. Some participants have provided calculations while others are clarifying assumptions about the arrangement of drivers and passengers. There is no explicit consensus yet on the correct method or final answer.
Participants are grappling with the constraints of the problem, particularly the limited number of licensed drivers and the total number of people needing to be seated. There is also confusion regarding how to properly account for the arrangements of drivers and passengers in the cars.
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.brendan said: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.
(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?
If you say "and then take the first two to be drivers of the two cars (in order)" then, yes, that is correct.brendan said:Just been looking at this problem again.
If there are 3 licensed drivers isn't there 3! (six times) they can be arranged ?
Be careful. There are not three cars so one of the people with a license has to be included amoung the 7 passengers.Which would leave 6 pasengers arranged into two cars say
You are assuming that the licensed person who is not driving will be in car A. That is not given.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