SOLUTION: There are 9 people lining up today to pay telephone bills. If three persons don:t want to follow each other, how many possible line-ups is possible?

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Question 1151217: There are 9 people lining up today to pay telephone bills. If three persons don:t want to follow each other, how many possible line-ups is possible?
Found 2 solutions by greenestamps, ikleyn:
Answer by greenestamps(13200)   (Show Source): You can put this solution on YOUR website!


Not clear....

"If three persons don't want to follow each other..."

is open to many different interpretations.


Answer by ikleyn(52781)   (Show Source): You can put this solution on YOUR website!
.

The total number of all possible line-ups is equal to the number of all permutations of 9 objects. i.e. 9! = 9*8*7*6*5*4*3*2*1.



From this number, you should subtract the number of all line-ups, where these 3 persons follow each other.


The number of such line-ups is  7*6!*3!= 7*(6*5*4*3*2*1)*(3*2*1).




The very first factor of 7 accounts for any of 7 possible positions of this triple in the line.


The factor 6! accounts for all possible permutations of remaining 9-3 = 6 persons.


Finally, the factor 3! accounts for all internal permutations inside this triple.


Thus the final answer is


    9*8*7*6*5*4*3*2*1 - 7*(6*5*4*3*2*1)*(3*2*1) = 332640.    ANSWER

Solved.


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