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?

- mathmath333

- chestercat

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- 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

so how many total stops are there?

- mathmath333

6

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## More answers

- ganeshie8

Careful, amsterdam is also a stop, the final stop

- mathmath333

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

- ganeshie8

there is no stealth mode
must be some bug haha

- mathmath333

yea it is

- mathmath333

now i can see u

- ganeshie8

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

- mathmath333

yes

- ganeshie8

|dw:1440807869323:dw|

- ganeshie8

How many choices are there for passenger1 ?

- mathmath333

6 choices

- ganeshie8

|dw:1440808107900:dw|

- ganeshie8

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

- mathmath333

5

- ganeshie8

right, see if you can finish it off

- mathmath333

6*5*4=120 ?

- ganeshie8

Yep!

- mathmath333

answer given is 455 way too far.

- ganeshie8

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

- ganeshie8

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

- 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

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

- ganeshie8

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

- mathmate

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

- mathmate

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

- 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

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

- 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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