2
$\begingroup$

Consider a hypothetical example.

There are $12$ tickets for a documentary film and exactly $12$ people to buy the tickets (one person can buy only one ticket).

There are $3$ ticketbooths for selling the tickets and each ticketbooth will sell exactly $4$ tickets.

Suppose the researcher labels the $12$ people with ID $1,2,...,12$ according to their arrivals.

Say, the first arrival buys his/her ticket from one of the $3$ ticketbooths.

Then, the second arrival buys his/her ticket from one of the $3$ ticketbooths. The second arrival can buy his/her ticket from the same ticket booth that the first arrival had bought or from one of the other two ticketbooths.

In this way, the last arrival buys his/her ticket.

The researcher has recorded from which ticketbooth which arrivals have bought the tickets.

Suppose from ticketbooth A, arrival #3, #5, #6, #12 have bought the tickets.

From ticketbooth B, arrival #1, #2, #9, #11 have bought the tickets.

From ticketbooth C, arrival #4, #7, #8, #10 have bought the tickets.

In how many ways the $12$ people can buy tickets from the $3$ ticketbooths?

If inside a ticketbooth the order of the ID doesn't matter, then the number of ways the $12$ people can buy tickets from the $3$ ticketbooths is $\frac{12!}{4!4!4!}.$

But for my example, since inside a ticketbooth it is naturally ordered (that is, the first arrival comes before the second arrival and so on), should I consider the order inside a ticketbooth? If so, then is the number of ways the $12$ people can buy tickets from the $3$ ticketbooths $(12!)$?

For the above example, is ORDER more meaningful or NOT?

EDIT:

For a reason, the researcher will take an expensive IQ test from only the fastest arrival of each ticketbooth. For our given example, the researcher will take the IQ test from the arrival #3 from ticketbooth A, arrival #1 from ticketbooth B, and arrival #4 from ticketbooth C. Now should I consider the order inside a ticketbooth or not to find the number of ways the $12$ people can buy tickets from the $3$ ticketbooths?

  • 1
    Both are valid interpretations of 'the number of ways the people can buy the tickets'. You can't really say one interpretation is better than the other, though one might be appropriate for computing the answer to specific questions. 'Does the order matter or not?' would have to be specified or implied by the problem. This is why it's important to word combinatorics/probability problems very carefully. – 2017-01-21
  • 0
    I wouldn't think it meaningful. I wouldn't consider (1,2,9,11) at a booth any different than (2,1,9,11). – 2017-01-21
  • 0
    Your edit misses the point. You haven't said what question the researcher is trying to answer, so it is still not clear whether order matters or not. Whether order matters should be stated by the person who asks the question. Sometimes it is obvious from the question, sometimes not. – 2017-01-21
  • 0
    @RossMillikan The researcher will calculate the probability that the $i$th arrival experiences the IQ test. So I need to know `In how many ways the 12 people can buy tickets from the 3 ticketbooths?` – 2017-01-21
  • 0
    Probably now you only care about which people get the test, so I would say it is just $12 \choose 3$ ways to select the test recipients and not worry about the ticket booths and how many ways people can visit them. It seems you are looking for a rule that says whether to consider different orders as different events. That is not a mathematical question. If you know people $1,2,9,11$ visit one booth, we can say that if order doesn't matter there is one way, while if order matters there are $4!$ ways. We can't say whether order matters to you. – 2017-01-21
  • 0
    Inside a booth whether it is arrival $(3,5,6,12)$ or $(5,12,3,6)$, I am going to take the IQ test from only the arrival number $3$. Now it seems order doesn't matter here. – 2017-01-21
  • 0
    And since inside a ticketbooth the order of the ID doesn't matter, then the number of ways the $12$ people can buy the tickets from the $3$ ticketbooths is $\frac{12!}{4!4!4!}.$ – 2017-01-21

0 Answers 0