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
A permutation refers to the different ways in which a set of objects can be arranged or ordered. In the context of seat arrangements, it refers to the different ways in which a group of people can be seated in a row or around a table.
The number of possible permutations can be calculated using the formula n!/(n-r)!, where n represents the total number of seats and r represents the number of people to be seated. For example, if there are 6 seats and 4 people, the number of possible permutations would be 6!/(6-4)! = 6!/2! = 6*5*4*3 = 360.
No, each permutation must be unique. This means that no two people can have the same seat in each permutation. For example, if there are 4 people and 4 seats, the permutations would be ABCD, ABDC, ACBD, etc. Repeating a permutation, such as AABB, would not be allowed.
A permutation refers to the different ways in which a set of objects can be arranged or ordered, while a combination refers to the different ways in which a subset of objects can be selected from a larger set. In seat arrangements, a permutation would be the different ways in which a group of people can be seated, while a combination would be the different ways in which a subset of those people can be chosen to sit in a specific order.
Permutations are helpful in various real life situations, such as in seating arrangements for events, assigning tasks to a group of people, and organizing schedules for a team. They can also be used in mathematical and statistical problems, as well as in computer programming.