SOLUTION: A class has 7 students. Two of the students, Rachel and Jonathan cannot stop bugging each other. How many line-ups are possible in which Jonathan and Rachel are not beside each oth

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Question 536053: A class has 7 students. Two of the students, Rachel and Jonathan cannot stop bugging each other. How many line-ups are possible in which Jonathan and Rachel are not beside each other?
Answer by Edwin McCravy(20055)   (Show Source): You can put this solution on YOUR website!
Suppose the set of students is {A,B,C,D,E,R,J}, where R is Rachel and J is Jonathan.

First calculate how many ways there are to arrange the seven "things",
{A,B,C,D,E,R,J}, whether R and J are together or not:

That is 7! or 5040

Now we must subtract from that all the ways R and J are together.
These consist of: 

(A)   the number of ways to arrange the 6 "things of the set {A,B,C,D,E,(RJ)},
with (RJ) considered as a SINGLE "thing", where R is on the left of J:

That is 6! or 720

(B)  the number of ways to arrange the 6 "things of the set {A,B,C,D,E,(JR)},
with (JR) considered as a SINGLE "thing", where J on the left of R:

That is also 6! or 720

So the final answer is 

7! - 6! - 6! = 7! - 2·6! = 5040 - 2(720) = 5040 - 1440 = 3600

Edwin
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