SOLUTION: How many ways can 7 people be seated in a row of chairs if two of the people, Wilma and Paul, refuse to sit each other?

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Question 1141007: How many ways can 7 people be seated in a row of chairs if two of the people, Wilma and Paul, refuse to sit each other?
Answer by ikleyn(52781) About Me  (Show Source):
You can put this solution on YOUR website!
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The number of all possible "in line" arrangements for 7 people is 

    7! = 1*2*3*4*5*6*7 = 5040

if NO RESTRICTIONS are imposed.



The number of all possible arrangements where Wilma and Paul are neighbors is

    2*6! = 2*(1*2*3*4*5*6) = 2*720 = 1440.


(The factor "2" in this formula reflects two possible arrangements of Wilma and Paul as neighbors.)



The number which is the answer to the problem's question is the difference


    7! - 2*6! = 5040 - 1440 = 3600.    ANSWER