# Permutations seat arrangements

• brendan
In summary, there are 210 different ways for 9 people to be seated into 2 cars, with only 3 people having licenses. This is calculated by choosing 2 of the 3 licensed drivers to be the drivers, and then arranging the remaining 7 passengers in the cars. This can be represented by 3! (6C3) + 3!(6C3) or 35 X 6 = 210.
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?

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.
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?

Last edited by a moderator:

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?

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

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 ?
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

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

## 1. What is a permutation in the context of seat arrangements?

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.

## 2. How do you calculate the number of possible permutations for a given number of seats and people?

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.

## 3. Can you have repeated permutations in seat arrangements?

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.

## 4. What is the difference between a permutation and a combination in seat arrangements?

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.

## 5. How are permutations helpful in real life situations?

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.

• Calculus and Beyond Homework Help
Replies
8
Views
3K
• Calculus and Beyond Homework Help
Replies
18
Views
1K
• Calculus and Beyond Homework Help
Replies
1
Views
667
• Precalculus Mathematics Homework Help
Replies
24
Views
2K
• Calculus and Beyond Homework Help
Replies
5
Views
1K
• Calculus and Beyond Homework Help
Replies
5
Views
1K
• Calculus and Beyond Homework Help
Replies
12
Views
1K
• General Math
Replies
3
Views
864
• Calculus and Beyond Homework Help
Replies
4
Views
1K
• Calculus and Beyond Homework Help
Replies
8
Views
1K