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very hard question

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very hard question

by bupbebeo » Fri Mar 26, 2010 7:58 am
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 book's answer is 360

could any one can help me why the answer.
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Source: — Problem Solving |

by thephoenix » Fri Mar 26, 2010 8:08 am
condition frankie wants to stand behind joe
total positon six from 1,2,3,4,5,6
to meet the condition frankie can take any position from 2 to 5(becoz when frankie takes 1 joe will be behind him so violating the condition)
case 1 when frankie is in 6th postion joe can take any one of available 5 position in 5c1 ways and rest 4 in 4! ways ; hence total ways=5*4!=120
case 2 frankie at 5th position # of ways=4*4!=96
like wise
for frankie at 4th position # of ways=3*4!=72
frankie at 3rd position # of ways=2*4!=48
frankie at 2nd position # of ways=1*4!=24
tot=120+96+72+48+24=36o
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by kevincanspain » Fri Mar 26, 2010 8:34 am
These 6 mobsters can arrange themselves in 6! ways. In exactly half of them, Frankie will be somewhere behind Joey, so the answer is 6!/2 = 360.

For every arrangement of the six people in which Joey is ahead of Frankie, switch Joey and Frankie and you get an otherwise identical arrangement in which Joey is behind Frankie.
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