Question

Ten people are going on a road trip and four of them decided to bring their cars. In how many ways can they be transported if we strictly abide by all of the following conditions (1) the owner will drive his own car (2) each car can hold a maximum of 5 passengers including the driver and (3) each car must transport at least 2 passengers

Answer 1

this sounds brutal and hard

Answer 2

I came up with 1,560, but I sort of rushed through the problem. 4 x 6C3 x 3C1 x 2C1 x 1C1 + 6 x 6C2 x 4C2 x 2C1 x 1C1 = 1,560.

Answer 3

why 6C3?

Answer 4

i see that 6 is 10 - 4...what is the 3?

Answer 5

Well, there are two basic scenarios. One of the 4 cars will have 4 peopleother3 cars having 2 apiece. Other scenario: 2 cars will have 3 people with the other 2 cars having 2 people. In thefirst scenario, the driver (driver having 4 in the car including himself) can pick 3 from the remaining 6. Once he does that, only 3 people left. This is for the first scenario.

Answer 6

The initial factor of 4 is from any of the cars could be the car with 4.

Answer 7

wait go slow....

Answer 8

In the second scenario, there are 6 ways to have 2 cars with 3 and 2 with 2. Okay, I'll go slow. I'll stop with what I wrote just now.

Answer 9

what is the first 4 for then?

Answer 10

shouldn't it be 3 * 6C3 since only 3 cars have that combination?

Answer 11

Could have 4222 or 2422 or 2242 or 2224. So, each group of numbers is the car configuration.

Answer 12

but order doesn't matter right?

Answer 13

i don't get what you meant by that

Answer 14

If we just concentrate on 4222 for the moment, 4222 staarts out as 1111 with drivers pre-assigned. So, there are 6 to choose from of which the car with 4 needs 3 more.

Answer 15

wait...what does 1111 mean?

Answer 16

each car have one passenger each?

Answer 17

So, in 4222, starting out as 1111 (drivers pre-assigned as owners of his own car), there are 6 people unassigned. So, the car which will have 4 needs to pick up 3 more from 6.

Answer 18

Yes, there must bst 2 in a car from the problem requirement.

Answer 19

okay..i get the first part of the sum...i don't get the second part though

Answer 20

why 6 and not 4 anymore?

Answer 21

Second part meaning the second scenario?

Answer 22

yes

Answer 23

Okay now we have 2 cars with 3 people (including driver) and 2 cars with 2 people. 4C2 ways or 6 to do that.

Answer 24

oh. shortcut

Answer 25

So, we have a factor of 6 and we can look at just 3322 (which again starts out as 1111). For the first car with 3, 6C2, then 4 left, so 4C2, 2 left, so 2C1

Answer 26

No other scenarios. Saying that a car can transport 5 people is just throwing a curve.

Answer 27

If we get 5 in a car, then there aren't enough people left to get at least 2 in a car.

Answer 28

I might have made a mistake somewhere as I said I rushed it, but most if not all of the logic is sound.

Answer 29

wait...go slow...

Answer 30

ok, I'll just answer questions for a while. We can go to any part of the problem or answer.

Answer 31

where is 4C2 applied?

Answer 32

in the car with 3 passengers?

Answer 33

My suggestion is to get an idea of the total (I say 2) # of scenarios and then break it down from there. 4C2 is in the second scenario. In any of the 6 - 3322 arrangements (3322, 3232, 3223, 2332, 2323, 2233), which start out as 1111 (they all do), there are 6 people who don't drive. The first car which will eventually have 3 people (already has 1 driver to start) has to pick 2 people. So, 6C2. Now there are 4 people left for the second car needing 2 more people (3 in all) to pick. So, 4C2.

Answer 34

wait nevermind. i get it already. thanks

Answer 35

Do you have the right answer from the folks who posed the problem? I would like to know if 1560 is right.

Answer 36

yes 1560 is right

Answer 37

I usually don't hurry through problems, but I really like combinations and permutations. So, I got in the groove.

Answer 38

it's tricky though. that might be what you like

Answer 39

Yes, I do like the tricky parts. It is a tricky problem in all because the drivers driving their own cars makes the cars unique.

Answer 40

i think i'm finally getting the hang of it

Answer 41

I don't know if I did the best job of explaining, because it's complicated. But you're no rookie, and you will understand where 95% of the people here will not.

Answer 42

in permutation/combination...im a complete rookie

Answer 43

I don't know how long I'll be here tonight, but tomorrow I can log on and see if you had any lingering questions. I'll stick around for a short while now though.

Answer 44

too bad the next question i posted was discrete mathematics

Answer 45

Completely off the topic of our question at hand: It would seem that one's Smartscore can be boosted also by asking questions. Is that the way it is? It seems like the Smartscore is a composite of all three.

Answer 46

yes smartscore can be boosted by asking questions. I used to be the user with the highest engagement score but sadly, i got dethroned.

Answer 47

I asked one question so far. No one could or would help me. I'm not blaming anyone at all. So, I stuck with my question and eventually found my own answer.

Answer 48

it happens

Answer 49

Nice talking with you and good luck with probability. If I'm around, I'll try to help out. Bye for now.