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Mobsters

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by gautamberry » Sat Aug 23, 2008 5:39 am
Please solve and explain this one.....came across on Manhattan
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Source: — Problem Solving |

Re: Mobsters

by sudhir3127 » Sat Aug 23, 2008 7:28 am
gautamberry wrote:Please solve and explain this one.....came across on Manhattan
i go with 360


here it goes...

6 people can arrange in 6! ways = 720 ways.
now

there's a 50% chance Frankie will be ahead of Joey in line, and a 50% chance Frankie will be behind Joey in line hence its

1/2*6! = 360.

hope it helps..
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by parallel_chase » Sat Aug 23, 2008 7:54 am
I think Sudhir's method is fastest,

This method will actually help you in understanding why it is 50%

You have to fix Joey in all the possible positions and calculate the possible arrangements for the remaining people.

1st: Joey is on the first position
Therefore no matter what is the position of Frankie he'll be behind Joey. Since Joey is fixed, the remaining five have 5! ways of being arranged.

2nd: Joey is in the second position
Frankie must be in position 3, 4, 5 or 6, meaning 4 possible positions. The remaining 4 people must be arranged in the remaining 4 positions. So 4*4!

3rd: Joey is in third position

Frankie must be in position 4, 5 or 6, meaning 3 possible positions. The remaining 4 people must be arranges in the remaining 4 positions. So, 3*4!

4th: Joey is in fourth position

Frankie must be in position 5 or 6, 2 possible positions. The remaining 4 people can be arranged in the 4 remaining positions. So 2*4!

5th: Joey is in fifth position

Frankie must be in position no. 6, one possible position. The remaining 4 people can be arranged in the remaining 4 positions, so 4!

Joey cannot be on position no. 6 because this is the last position, meaning that Frankie could not be behind him.

So, 5! + 4*4! + 3*4! + 2*4! + 4! = 360

Hope it helps.
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by gautamberry » Sat Aug 23, 2008 12:04 pm
thanx guys.....both methods very helpful..........
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