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by fangtray » Tue Sep 20, 2011 5:15 am
Six mobsters have arrived at the theater for the premiere of the film "Goodbuddies." One of the mobsters, Frankie, is an informer, and he's afraid that another member of his crew, Joey, is on to him. Frankie, wanting to keep Joey in his sights, insists upon standing behind Joey in line at the concession stand, though not necessarily right behind him. How many ways can the six arrange themselves in line such that Frankie's requirement is satisfied?
6
24
120
360
720

Ignoring Frankie's requirement for a moment, observe that the six mobsters can be arranged 6! or 6 x 5 x 4 x 3 x 2 x 1 = 720 different ways in the concession stand line. In each of those 720 arrangements, Frankie must be either ahead of or behind Joey. Logically, since the combinations favor neither Frankie nor Joey, each would be behind the other in precisely half of the arrangements. Therefore, in order to satisfy Frankie's requirement, the six mobsters could be arranged in 720/2 = 360 different ways.

The correct answer is D.

the answer explanation says frankie must be either ahead of or behind joey, but the question stem says frankie wants to be behind joey. is there something i am missing?
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Source: — Problem Solving |

by sl750 » Tue Sep 20, 2011 5:47 am
In the 720 possible scenarios, Frankie could be either ahead or behind Joey. So in half the cases, he ends up being behind him. That is what it means
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by Brent@GMATPrepNow » Tue Sep 20, 2011 5:52 am
Or we can say: for every arrangement where Frankie is ahead of Joey, there is an arrangement where Joey is ahead of Frankie.

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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