Ann, Mark, Dave and Paula line up at a ticket window. In how many way

Take the task of arranging the 4 people and break it into stages.

We’ll begin with the most restrictive stage.

Stage 1: Select a place in line for Dave
Since Dave must be in the 3rd position, we can complete stage 1 in 1 way

Stage 2: Select a place in line for Ann
Now that Dave is positioned in line, there are 3 available places remaining.
So we can complete stage 2 in 3 ways

Stage 3: Select a place in line for Mark
Now that Dave and Ann are positioned in line, there are 2 available places remaining.
So we can complete stage 3 in 2 ways

Stage 4: Select a place in line for Paula
Now that Mark, Dave and Ann are positioned in line, there is 1 space remaining.
So we can complete stage 4 in 1 way

By the Fundamental Counting Principle (FCP), we can complete all 4 stages (and thus arrange all 4 people) in (1)(3)(2)(1) ways (= 6 ways)

Answer: D

Note: the FCP can be used to solve the MAJORITY of counting questions on the GMAT. So, be sure to learn it.

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\(?\,\,\,:\,\,\,{\text{# }}\,\,{\text{ways}}\,\,{\text{with}}\,\,{\text{Dave position}}\,\,{\text{fixed}}\)

We may consider Dave does not "exist": put Ann, Mark and Paula in line (at the 3 positions available to them)!

\(? = {P_3} = 3! = 6\)


This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.

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