Math questions from any Manhattan Prep GMAT Computer Adaptive Test.
denise
 
 

mobster question (CAT2)

by denise Sat Jan 24, 2009 9:35 pm

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?

The answer posted for this question is
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.

Is there another way to show mathematically how to derive this answer? The way the answer is laid out does not really make sense to me...thanks!
experts
 
 

Explanation

by experts Sun Jan 25, 2009 1:15 am

Explanation:
Suppose J is Joey and F is Frankie; we can say the line is as follows
= > 1 2 3 4 5 6
Now break the problem into 5 steps; where F is behind J.

1) = > J 2 3 4 5 6 ;here J is at 1, and F can stand at any place ( 2,3,4,5,6) so 5 ways ; and other 4 people can be arranged in 4! ways so total = 4! * 5
2) = > 1 J 3 4 5 6 ;here J is at 2, and F can stand at any place ( 3,4,5,6) so 4 ways ; and other 4 people can be arranged in 4! ways so total = 4! * 4
3) = > 1 2 J 4 5 6 ;here J is at 3, and F can stand at any place ( 4,5,6) so 3 ways ; and other 4 people can be arranged in 4! ways so total = 4! * 3
4) = > 1 2 3 J 5 6 ;here J is at 4, and F can stand at any place ( 5,6) so 2 ways ; and other 4 people can be arranged in 4! ways so total = 4! * 2
5) = > 1 2 3 4 J 6 ;here J is at 5, and F can stand at any place ( 6) so 1 way ; and other 4 people can be arranged in 4! ways so total = 4! * 1

so total ways are = (4! * 5) + (4! * 4) + (4! * 3) + (4! * 2) + (4! * 1) = 4! * 15 = 360
JonathanSchneider
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Re: mobster question (CAT2)

by JonathanSchneider Thu Feb 19, 2009 2:25 pm

Nice work on the above.

As to the original solution, this is a great case of radically simplifying an otherwise complex problem. While the above solution works, it is certainly harder than seeing that there is a 50% chance that J will be behind F. Why 50%? Because J and F have the same chances for being in any given seat. There is nothing that says J will be behind F more often than F is behind J. Why 720? This = 6!, which is just the number of ways that we can arrange six things, or, in this case, the number of ways that we could line up the six mobsters.