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combinations

by simba12123 » Tue Nov 04, 2008 2:36 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?


6
24
120
360
720

qa is 360

source mgmat

I cant really understand why we are dividing the total amount of possibilities by 2.
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by dmateer25 » Tue Nov 04, 2008 3:15 pm
Here are the possibilities:

FJ----
F-J---
F--J--
F---J-
F----J
-FJ---
-F-J--
-F--J-
-F---J
--FJ--
--F-J-
--F--J
---FJ-
---F-J
----FJ

So there are 15 ways that Joey can be behind Frankie.

Now the other 4 mobsters can be sorted in any of the spots for each of the 15 possibilities:

4! = 24

15 * 24 = 360
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by uttara » Wed Nov 05, 2008 10:42 am
total arrangements possible = 6!=720

Now in exaclty 50% of these J will be before F, hence
required probablity = 1/2*720 = 360
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by logitech » Wed Nov 05, 2008 10:55 am
uttara wrote:total arrangements possible = 6!=720

Now in exaclty 50% of these J will be before F, hence
required probablity = 1/2*720 = 360
How come 50% of these J will be before F ? Lets hear some explanations
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by dmateer25 » Wed Nov 05, 2008 12:39 pm
logitech wrote:
uttara wrote:total arrangements possible = 6!=720

Now in exaclty 50% of these J will be before F, hence
required probablity = 1/2*720 = 360
How come 50% of these J will be before F ? Lets hear some explanations
I did my F's and J's backward in my answer, but if you think about it if Joey can be in front of Frankie 15 ways then Frankie can also be in front of Joey 15 ways.

So 50% of the total 720 Joey is in front of Frankie and 50% Frankie is in front of Joey.
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