mathmath333 one year ago A Malaysia Airlines is going from kuala lumpur to amsterdam and stops in dubai,jerusalem,sofia,rome and vienna on the way. 3 passangers enter the Airplane with 3-different tickets. How many different collection of tickets may they have?

1. mathmath333

\large \color{black}{\begin{align} & \normalsize \text{A Malaysia Airlines is going from kuala lumpur to amsterdam and stops } \hspace{.33em}\\~\\ & \normalsize \text{ in dubai,jerusalem,sofia,rome and vienna on the way.} \hspace{.33em}\\~\\ & \normalsize \text{3 passangers enter the Airplane with 3-different tickets.} \hspace{.33em}\\~\\ & \normalsize \text{ How many different collection of tickets may they have?.} \hspace{.33em}\\~\\ \end{align}}

2. ganeshie8

so how many total stops are there?

3. mathmath333

6

4. ganeshie8

Careful, amsterdam is also a stop, the final stop

5. mathmath333

i cannot see ur logo r u in invisible (stealth mode)

6. ganeshie8

there is no stealth mode must be some bug haha

7. mathmath333

yea it is

8. mathmath333

now i can see u

9. ganeshie8

good, so there are 6 stops and we're told that the 3 passengers are carrying different tickets

10. mathmath333

yes

11. ganeshie8

|dw:1440807869323:dw|

12. ganeshie8

How many choices are there for passenger1 ?

13. mathmath333

6 choices

14. ganeshie8

|dw:1440808107900:dw|

15. ganeshie8

after that, how many choices are there for the ticket of passenger2 ?

16. mathmath333

5

17. ganeshie8

right, see if you can finish it off

18. mathmath333

6*5*4=120 ?

19. ganeshie8

Yep!

20. mathmath333

answer given is 455 way too far.

21. ganeshie8

Perhaps it has something to do with $$\dbinom{15}{3}$$ notice that $$15=5+4+3+2+1$$

22. ganeshie8

if we consider the case that not all 3 passengers enter at the same start point

23. mathmate

Strange, I get 1330. Starting from Kuala Lumpur, there are 6 destinations, and progressively reduced for passengers embarking at later airports. First passenger can have a choice of 1+2+3+4+5+6=21 tickets. Second ha 20, and third 19. So, assuming name of passengers do not count (as different), we have 21*20*19/6=1330 possible collection of tickets.

24. ganeshie8

Exactly, looks the passengers only enter at the 5 intermediate stops

25. ganeshie8

no doubt question is ambiguous, as it stands, 1330 looks more correct to me !

26. mathmate

Ah, that's why, so 15*14*13/6=455, there we go! Thanks!

27. mathmate

"...on the way.3 passangers enter the Airplane with 3-different tickets." But there is a stop after "on the way"!

28. ganeshie8

thats good catch! probably it should read \large \color{black}{\begin{align} & \normalsize \text{A Malaysia Airlines is going from kuala lumpur to amsterdam and stops } \hspace{.33em}\\~\\ & \normalsize \text{ in dubai,jerusalem,sofia,rome and vienna. On the way,} \hspace{.33em}\\~\\ & \normalsize \text{3 passangers enter the Airplane with 3-different tickets.} \hspace{.33em}\\~\\ & \normalsize \text{ How many different collection of tickets may they have?.} \hspace{.33em}\\~\\ \end{align}}

29. mathmate

@ganeshie8 \$455 would be correct for your corrected version of the problem! lol

30. phi

I was puzzled. the first interpretation, where they can get on kuala lumpur there are 7C2 trips (pick two cities, order does not matter) and number of combinations is (7C2) C 3 = 1330 but if they get on after kuala lumpur it is (6C2) C 3 = 15C3 = 455