Permutations and Combinations people and cars

[brendan]

Homework Statement

A family of nine has two vehicles, each of which can hold a maximum of five people.

Homework Equations

How many different ways are there of allocating people to the cars?

If three of the memebers of the family only have a drivers licence, how many different ways are there of allocating people to the cars?

The Attempt at a Solution

Question part a.

9 people 3 cars

9! / (6! * 3!) = 84


Question part b.


9! / 6! = 504




Are these answers on track?

 

[tiny-tim]

Hi brendan!

A family of nine has two vehicles, each of which can hold a maximum of five people.
…
How many different ways are there of allocating people to the cars?

If three of the memebers of the family only have a drivers licence, how many different ways are there of allocating people to the cars?
…
Question part a.

9 people 3 cars

9! / (6! * 3!) = 84

erm … two cars!

but it's the wrong method anyway …

try again, using the 4 or the 5

Question part b.

9! / 6! = 504

Nooo … first fill up the drivers' seats, then fill up the rest.

 

[brendan]

In The formula


n! / (n-r)! would the n be the number of seats or the number of people?

There are 10 seats between two cars and 9 people.

would it be

10! / (10-9)!
= 3628800


or

10! / (10-5)! X 9! / (10-4)!

= 152409600

regards

 

[tiny-tim]

hi brendan!

you're just guessing, aren't you?

sorry, but you really need to read this up again, right from the start

to help you along, it may be easier itf you use the ⁿC_m notation (pronounced "n choose m") …

the advantage is that that reminds you that it is the number of ways of choosing m items from n …

in this case, you can only split the passengers 4,5 or 5,4

so you need to choose 4 (or 5) to go in the first car …

read the chapter again, and then come back and have another try

 

[brendan]

Thanks mate.

I read this as

9! / (9-5)! X 9! / (9-4)!

9C5 x 9C4

That is choose how many ways 5 people from the 9 people can go into the first car and ways 4 from the 9 people to go into the second car.

Now order is not important so we use the formula

n!/ (n-r! * r!)

which is for first car

9!/ (9-5! * 5!) and second car 9!/ (9-4! * 4!)

= 126 and 126

Do we than multiply together to get

15876 ?

regards
Brendan.

 

[brendan]

Sorry we should just add them

so 252 combinations

It gets pretty tricky with multiple combinations

 

[Dick]

Yes, it does get tricky. But the reason you add them is because you can either put five people into the first car or five people into the second. Those are your only options.

 

[tiny-tim]

Hi brendan!

… 9C5 x 9C4

That is choose how many ways 5 people from the 9 people can go into the first car and ways 4 from the 9 people to go into the second car.

Now order is not important so we use the formula

n!/ (n-r! * r!)

which is for first car

9!/ (9-5! * 5!) and second car 9!/ (9-4! * 4!)

= 126 and 126

Sorry we should just add them

so 252 combinations

Yes, it does get tricky. But the reason you add them is because you can either put five people into the first car or five people into the second. Those are your only options.

brendan, it helps if you practice different ways of counting the same things …

in this case, you can either say (as you did) choose how many ways 5 people from the 9 people can go into the first car and ways 4 from the 9 people to go into the second car,

which is 9C5 + 9C4

or you can say we need to put 5 pepople in one of the cars, and it can be either the first car or the second car, so choose how many ways 5 people from the 9 people can go into the first car and ways 5 from the 9 people to go into the second car,

which is 9C5 + 9C5

or you can say (… what do you think? wink: …),

which is 9C4 + 9C4.

Now what about part b.? ​

 

[brendan]

For Part B.

We have 3 drivers so they can be arranged in 3! ways = 6.
So after the drivers are allocated we still have 7 people to allocate and they can be allocated in groups of 4 or 3 so 7C3 or 7C4 which is the same = 35

We than mutiply those 35 combinations by the number of driver combinations which is 6.


therefore 6 x 35

= 210

regards
Brendan

 

[tiny-tim]

For Part B.

We have 3 drivers so they can be arranged in 3! ways = 6.
So after the drivers are allocated we still have 7 people to allocate and they can be allocated in groups of 4 or 3 so 7C3 or 7C4 which is the same = 35

We than mutiply those 35 combinations by the number of driver combinations which is 6.


therefore 6 x 35

= 210

regards
Brendan

mmm … is it 35 or is it 70?