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

Roland2rule wrote: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.

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

Roland2rule wrote: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

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