Question

8 people, 2 cars (5 places each), everyone can drive. How many possibilities?

Answer 1

is this the full info given to u????

Answer 2

nd wht is your q'? be more pecific

Answer 3

q' ? How many ways i have to sit those 8 people like this.

What im thinking is 2!x [ 5! / (5-4)! ]

Answer 4

but i know there is a problem because the driver must be in the driver seat. you cant have 3 or 4 people in a car and none in the driver seat

Answer 5

I think its 8C5 for each car and then you have two cars, so its 2*8C5

Answer 6

or simply 10C8 ???

Answer 7

You cant use combinations because you don't have reposition. If you sit the first guy in the car you only have 7 left

Answer 8

Fixing the drivers: 1x1
now for the others you can have 4 ou each car or you can have 5 in one and 3 on the other

i think the final answer is

1x1( 6P3 x 3P3 + 2! x 6P4 x 2P2)

Am i correct?

Answer 9

@Mertsj ?

Answer 10

maybe for the drivers instead of 1x1 i should use 8P2...

Answer 11

Not 100% sure this covers every possibility...

First pick the two drivers, pick 2 from 8. Then from the 6 people who are left, pick 4 for one car (there's only four seats left), two for the the other car, and so on.

\[\Large (8C2*6C4*2C2) + (8C2*6C3*3C3) + (8C2*6C2*4C4 )\]

Answer 12

There might be an empty seat in each car. Does that matter?

Answer 13

You can have empty seats. Thats why im confused

Answer 14

Does the actual position of the people in the car matter? ie other than the driver? It doesn't mention that it matters which seat they're in.

Answer 15

8P2 x ( 6P3 x 3P3 + 2! x 6P4 x 2P2) = 120960

should be this?

Answer 16

The method I gave gives 1400 ways:
http://www.wolframalpha.com/input/?i=%288Choose2%29%E2%88%97%286Choose4%29%E2%88%97%282Choose2%29%2B%288Choose2%29%E2%88%97%286Choose3%29%E2%88%97%283Choose3%29%2B%288Choose2%29%E2%88%97%286Choose2%29%E2%88%97%284Choose4%29

Answer 17

403200

Answer 18

how?

Answer 19

"You can have empty seats. Thats why im confused"

It doesn't actually say it matters who sits in what seat... I think it only matters who is driving, and who is in the car. I've seen a similar problem recently, which was solved using combinations similar to how I solved it.

imo the only thing the problem is concerned with is who is in each car, and who is driving - not where they sit, empty seats etc.

Answer 20

teacher just gave me the answer. It was 182..
After some time i ended up realizing it was combinations indeed...


Here's what i have: 8C4 + 8C3 + 8C5 = 182

Simple as that... It doesnt even matter who were driving -.-
But thanks for your answer @agent0smith

Answer 21

Oh if it doesn't matter who's driving...

From 8 people pick 5 for the first car, then from the remaining 3 pick 3. From 8 pick 4 for the first, 4 for the second. From 8 pick 3 for the first, 5 for the second.

\[\Large (8C5*3C3) + (8C4*4C4) + (8C3*5C5 )\]
and all the 3C3, 4C4, 5C5 are 1 anyway... but I like showing them that way in the context of the question.

Answer 22

I guess it doesn't matter who you pick to drive, cos once the people are in the car, one of them can decide to drive.

Answer 23

yep.. Thats what i figured out. I think the main fault was from the problem itself. Wasn't clear enough

Answer 24

Yeah, i've seen almost an exact problem, but it didn't mention the driver.