Ann, Mark, Dave and Paula line up at a ticket window

[BTGmoderatorRO]

Ann, Mark, Dave and Paula line up at a ticket window. In how many ways can they arrange themselves so that Dave is third in line from the window?

a 24
b 12
c 9
d 6
e 3
OA is D
How can I permutate this? Pls and Expert should show me the exact logic behind this

[GMATGuruNY]

Ann, Mark, Dave and Paula line up at a ticket window. In how many ways can they arrange themselves so that Dave is third in line from the window?

a 24
b 12
c 9
d 6
e 3

Number of options for the third position = 1. (Must be Dave.)
Number of ways to arrange the remaining 3 people = 3!.
To combine the options above, we multiply:
1*3! = 6.

The correct answer is D.

Most of the answer choices are quite small, implying that we can simply list all of the arrangements in which D is in the third position:
AMDP
APDM
MADP
MPDA
PADM
PMDA.
Total ways = 6.

[[email protected]]

Hi Roland2rule,

We're told that Ann, Mark, Dave and Paula line up at a ticket window. We're asked for the number of ways that they can arrange themselves so that Dave is THIRD in line from the window. This question can be solved in a number of different ways. Mitch has shown a couple of really fast ways to get to the correct answer. You could also start with the overall total and eliminate options.

With 4 people, there are 4! = 24 different ways to arrange the people in line. Since each person has an equal possibility of being in any of the 4 spots, 'locking' Dave into the third spot means that 3/4 of HIS options are not possible - thus we can subtract 3/4 of 24 from that total:

24 - (3/4)(24) =
24 - 18 =
6

Final Answer: D

GMAT assassins aren't born, they're made,
Rich

[Jeff@TargetTestPrep]

Ann, Mark, Dave and Paula line up at a ticket window. In how many ways can they arrange themselves so that Dave is third in line from the window?

a 24
b 12
c 9
d 6
e 3

Since Dave MUST BE third in line, we only care about arranging the other 3 people and they can be arranged in 3! = 6 ways.

Answer: D