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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?

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

  2. ganeshie8
    • one year ago
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    so how many total stops are there?

  3. mathmath333
    • one year ago
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    6

  4. ganeshie8
    • one year ago
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    Careful, amsterdam is also a stop, the final stop

  5. mathmath333
    • one year ago
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    i cannot see ur logo r u in invisible (stealth mode)

  6. ganeshie8
    • one year ago
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    there is no stealth mode must be some bug haha

  7. mathmath333
    • one year ago
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    yea it is

  8. mathmath333
    • one year ago
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    now i can see u

  9. ganeshie8
    • one year ago
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    good, so there are 6 stops and we're told that the 3 passengers are carrying different tickets

  10. mathmath333
    • one year ago
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    yes

  11. ganeshie8
    • one year ago
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    |dw:1440807869323:dw|

  12. ganeshie8
    • one year ago
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    How many choices are there for passenger1 ?

  13. mathmath333
    • one year ago
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    6 choices

  14. ganeshie8
    • one year ago
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    |dw:1440808107900:dw|

  15. ganeshie8
    • one year ago
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    after that, how many choices are there for the ticket of passenger2 ?

  16. mathmath333
    • one year ago
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    5

  17. ganeshie8
    • one year ago
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    right, see if you can finish it off

  18. mathmath333
    • one year ago
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    6*5*4=120 ?

  19. ganeshie8
    • one year ago
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    Yep!

  20. mathmath333
    • one year ago
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    answer given is 455 way too far.

  21. ganeshie8
    • one year ago
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    Perhaps it has something to do with \(\dbinom{15}{3}\) notice that \(15=5+4+3+2+1\)

  22. ganeshie8
    • one year ago
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    if we consider the case that not all 3 passengers enter at the same start point

  23. mathmate
    • one year ago
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    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
    • one year ago
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    Exactly, looks the passengers only enter at the 5 intermediate stops

  25. ganeshie8
    • one year ago
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    no doubt question is ambiguous, as it stands, 1330 looks more correct to me !

  26. mathmate
    • one year ago
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    Ah, that's why, so 15*14*13/6=455, there we go! Thanks!

  27. mathmate
    • one year ago
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    "...on the way.3 passangers enter the Airplane with 3-different tickets." But there is a stop after "on the way"!

  28. ganeshie8
    • one year ago
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    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
    • one year ago
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    @ganeshie8 $455 would be correct for your corrected version of the problem! lol

  30. phi
    • one year ago
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    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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