# Permutations and Combinations

1. Feb 18, 2009

### brendan

1. The problem statement, all variables and given/known data

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

2. Relevant 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?

3. The attempt at a solution

Question part a.

9 people 3 cars

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

Question part b.

9! / 6! = 504

2. Feb 18, 2009

### tiny-tim

Hi brendan!
erm two cars!

but it's the wrong method anyway …

try again, using the 4 or the 5
Nooo … first fill up the drivers' seats, then fill up the rest.

3. Feb 18, 2009

### 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

4. Feb 18, 2009

### 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 nCm 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

5. Feb 18, 2009

### brendan

Thanks mate.

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.

6. Feb 18, 2009

### brendan

Sorry we should just add them

so 252 combinations

It gets pretty tricky with multiple combinations

7. Feb 18, 2009

### 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.

8. Feb 19, 2009

### tiny-tim

Hi brendan!
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.

9. Mar 5, 2009

### 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

10. Mar 7, 2009

### tiny-tim

mmm … is it 35 or is it 70?