Question

There are 8 cars driving in a race. They are labeled with the letters A through H. No cars arrive at the exact same time. Car D comes in last place. Cars E and C arrive consecutively. Car B arrives after car F, but before car A. Car E arrives before car A. In how many different orders could the cars have finished the race?

I can see that 8! would be the correct answer of 8 cars without the additional data. But not sure how to answer it with the additional information. Thanks

Answer 1

The answer isn't necessarily 8.

Answer 2

Need some "scratch paper" to sketch this out a little...

Answer 3

D must be last
E & C arrive one after the other (so E can't be next to last because C would be last)

Answer 4

Ya... I was thinking the answer would use 8! (8 factoral) ... but not sure where to go from there. I have scratched papered it out 20 times and can't seem to get anywhere.

Answer 5

B arrives after F (so F can't be next to last)

etc etc.

Answer 6

You could, in theory, set up a full factorial sort of problem, but it would be complicated

Answer 7

maybe you can draw it out like those brain teaser games, and fill in the letters you know based on process of elimination, and also "x" out spaces that letters CAN'T be. Then write the factorial stuff on the only areas that you cannot determine for sure.

Answer 8

I'd be HAPPY for an elegant solution though... maybe one has arrived!

Answer 9

Jake... elegant solutions are always nice.

Answer 10

I know it... wish I had one!

Answer 11

that makes two of us... thanks for the effort.

Answer 12

Of all the full 8! orders, you can divide by 8 immediately, since you know the last place car

Answer 13

Jake... ahh... that is a clue.

Answer 14

the other constraints will leave you with several possibilities each for different areas of the order, so I suspect you will have some smaller "order" terms like 3! if you found a group of 3 cars that could finish in any order and not "break the rules" of the problem

Answer 15

Unfortunately Jake none of the cars can finish in any order because D is always last no matter what

Answer 16

yes, right.... so, max possible if no other rules is 7! with D last, but all those constraints "operate" in the space of the remaining 7 places

Answer 17

true... we know cars G and H can arrive any where besides last place

Answer 18

yes...

Answer 19

D can be only 1 place

1!*

E and C arive consecutively , E can be place in 6 places and so can C

Answer 20

well, if I could count to 8 it would help... :(

Answer 21

E can have 6 spots (1-6) and C can also, but they are (2-7)

Like @Outkast3r09 said...

Answer 22

Going to delete my counting failures... clean up the thread...

Answer 23

ok here is the break down of where people can arrive
A- 3,4,5,6,7
B-2,3,4,5,6,7
C-1,2,3,4,5,6,7
D-8
E-1,2,3,4,5,6,7
F-1,2,3,4,5,6,7
G-1,2,3,4,5,6,7
H-1,2,3,4,5,6,7

Answer 24

@Outkast3r09 thats good thinking...

Answer 25

A=5! [A cannot be 1st, 8th or 2nd due to E and C consecutively
D=1!
E=6!
C=6! there is a note after that A must always arrive after E so that means A can be 7 spots but E and C can only be put into 6 spots since E and C can be interchanged

Answer 26

B can be in 5 spots it can't be 1st,7th,or 8th

Answer 27

F can be any but 8th

Answer 28

and 7th

Answer 29

Is total number of ways cars can be arranged equal to a sum of these? That doesn't "feel" right, but I'm not seeing it yet.

Answer 30

thanks Jake and Outkast... working through these suggestions now

Answer 31

Outkast... does that leave G = 8! and H = 8I ? If so, do I add up 5!+1!+6!+6!+5!+6!+8!+8! ?

Answer 32

no it doesn't none of the cars can have more than 7

Answer 33

I'm a little concerned... often these order problems are set up with the number of options for each position (as a factorial) times the number of options for the next position (different factorial) and so on. It's not additive...

Answer 34

and most wont have 6

Answer 35

yeah it's multiplicative

Answer 36

I gotta run... thanks for jumping in to help... I wasn't getting coherent thought going quickly enough :) stuck in the brainstorming and scratch paper mode. :)

Answer 37

Outkast: so, if G = 7! and H = 7! ... where does that leave us? Are you saying I should multiply the individual results times each other?