mathmath333
  • mathmath333
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?
Mathematics
chestercat
  • chestercat
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mathmath333
  • mathmath333
\(\large \color{black}{\begin{align} & \normalsize \text{A Malaysia Airlines is going from kuala lumpur to amsterdam and stops } \hspace{.33em}\\~\\ & \normalsize \text{ in dubai,jerusalem,sofia,rome and vienna on the way.} \hspace{.33em}\\~\\ & \normalsize \text{3 passangers enter the Airplane with 3-different tickets.} \hspace{.33em}\\~\\ & \normalsize \text{ How many different collection of tickets may they have?.} \hspace{.33em}\\~\\ \end{align}}\)
ganeshie8
  • ganeshie8
so how many total stops are there?
mathmath333
  • mathmath333
6

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ganeshie8
  • ganeshie8
Careful, amsterdam is also a stop, the final stop
mathmath333
  • mathmath333
i cannot see ur logo r u in invisible (stealth mode)
ganeshie8
  • ganeshie8
there is no stealth mode must be some bug haha
mathmath333
  • mathmath333
yea it is
mathmath333
  • mathmath333
now i can see u
ganeshie8
  • ganeshie8
good, so there are 6 stops and we're told that the 3 passengers are carrying different tickets
mathmath333
  • mathmath333
yes
ganeshie8
  • ganeshie8
|dw:1440807869323:dw|
ganeshie8
  • ganeshie8
How many choices are there for passenger1 ?
mathmath333
  • mathmath333
6 choices
ganeshie8
  • ganeshie8
|dw:1440808107900:dw|
ganeshie8
  • ganeshie8
after that, how many choices are there for the ticket of passenger2 ?
mathmath333
  • mathmath333
5
ganeshie8
  • ganeshie8
right, see if you can finish it off
mathmath333
  • mathmath333
6*5*4=120 ?
ganeshie8
  • ganeshie8
Yep!
mathmath333
  • mathmath333
answer given is 455 way too far.
ganeshie8
  • ganeshie8
Perhaps it has something to do with \(\dbinom{15}{3}\) notice that \(15=5+4+3+2+1\)
ganeshie8
  • ganeshie8
if we consider the case that not all 3 passengers enter at the same start point
mathmate
  • 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.
ganeshie8
  • ganeshie8
Exactly, looks the passengers only enter at the 5 intermediate stops
ganeshie8
  • ganeshie8
no doubt question is ambiguous, as it stands, 1330 looks more correct to me !
mathmate
  • mathmate
Ah, that's why, so 15*14*13/6=455, there we go! Thanks!
mathmate
  • mathmate
"...on the way.3 passangers enter the Airplane with 3-different tickets." But there is a stop after "on the way"!
ganeshie8
  • ganeshie8
thats good catch! probably it should read \(\large \color{black}{\begin{align} & \normalsize \text{A Malaysia Airlines is going from kuala lumpur to amsterdam and stops } \hspace{.33em}\\~\\ & \normalsize \text{ in dubai,jerusalem,sofia,rome and vienna. On the way,} \hspace{.33em}\\~\\ & \normalsize \text{3 passangers enter the Airplane with 3-different tickets.} \hspace{.33em}\\~\\ & \normalsize \text{ How many different collection of tickets may they have?.} \hspace{.33em}\\~\\ \end{align}}\)
mathmate
  • mathmate
@ganeshie8 $455 would be correct for your corrected version of the problem! lol
phi
  • 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

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