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P & C problm

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by jogi1984 » Thu Sep 15, 2011 7:42 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
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by Anurag@Gurome » Thu Sep 15, 2011 8:51 pm
jogi1984 wrote: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
Solution:
The six mobsters can be arranged in 6! = 720 ways.
In half of these arrangements Joey will be before Frankie and in the other half Frankie will be before Joey.
So, required answer is 720/2 = 360
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by Juggernaut_86 » Thu Sep 15, 2011 9:02 pm
The 6 mobsters - ABCDJF (J-Joey, F-Frankie) - can be arranged in 6!=720 ways WITHOUT any constraints.

Frankie's requirement can be met in 5 ways -

1) Assuming that the order is - ABCDJF i.e. Joey's position is second to the last and Frankie's position is last, the six mobsters can be arranged in 5! ways.
We will have to consider J & F as one entity.

2) Assuming that the order is - ABCJDF, the six mobsters can be arranged in 4*4! ways (4! is for ABCD and the multiplier 4 is for the number of positions that J and F can take with 1 mobster between them).

3) Assuming that the order is - ABJCDF, the six mobsters can be arranged in 3*4! ways (4! is for ABCD and the multiplier 3 is for the number of positions that J and F can take with 2 mobsters between them).

4) Assuming that the order is - AJBCDF, the six mobsters can be arranged in 2*4! ways (4! is for ABCD and the multiplier 2 is for the number of positions that J and F can take with 3 mobsters between them).

5) Assuming that the order is - JABCDF, the six mobsters can be arranged in 1*4! ways (4! is for ABCD and the multiplier 1 is for the number of positions that J and F can take with 4 mobsters between them).

Adding the numbers obtained from these 5 scenarios, we get - 5! + 4*4! + 3*4! + 2*4! + 1*4! = 360


Now we need to subtract the 360 ways from the 720 "unrestricted" ways in which the 6 mobsters can be positioned.

Therefore the answer is D - 360.

Thanks!
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