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MGMAT Cat | Combinatorics

by [email protected] » Sat Oct 27, 2012 10:04 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
I am not able to understand the combinations to be subtracted. Would be helpful, if someone can present the solution with the Glue method described in Manhattan strategy guides?
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by Brent@GMATPrepNow » Sat Oct 27, 2012 11:00 pm
If we remove the restriction that Frankie must stand behind Joey, then the mobsters can be arranged in 6! (720) different ways.
Now, of these 720 different arrangements, it is clear that Frankie will be standing behind Joey half the time, and Joey will be standing behind Frankie half the time.

So, the answer is 720/2 [spoiler]= 360 = D[/spoiler].

Cheers,
Brent
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by [email protected] » Mon Oct 29, 2012 7:17 pm
Can you please elaborate on this.it is clear that Frankie will be standing behind Joey half the time, and Joey will be standing behind Frankie half the time.

I dont understand.

Thanks.
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by Brent@GMATPrepNow » Mon Oct 29, 2012 10:21 pm
[email protected] wrote:Can you please elaborate on this.it is clear that Frankie will be standing behind Joey half the time, and Joey will be standing behind Frankie half the time.

I dont understand.

Thanks.
For every arrangement where Frankie is behind Joey, we can create another arrangement by switching their places to have Joey behind Frankie.

Cheers,
Brent
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