SOLUTION: Seven students are asked to line up for a photo. If the students Jack and Jill are both in the photo but do not want to stand together, then the number of different line-ups possi

Algebra ->  Permutations -> SOLUTION: Seven students are asked to line up for a photo. If the students Jack and Jill are both in the photo but do not want to stand together, then the number of different line-ups possi      Log On


   

Question 851580: Seven students are asked to line up for a photo. If the students Jack and Jill are both in
the photo but do not want to stand together, then the number of different line-ups possible
is:
Answer by Edwin McCravy(20055) About Me  (Show Source):
You can put this solution on YOUR website!
Suppose the seven students are

Ann
Barbara
Carl
Doug
Edith
Jack
Jill

If it did not matter if Jack and Jill stood together,
the answer would by 7! = 5040 ways.

But since they cannot, we must subtract from that the number
of ways they could stand together.

That's either the number of permutations of these 6 "things":

Ann
Barbara
Carlhttp://www.algebra.com/tutors/linear.solver?solver_action=plug
Doug
Edith
Jack&Jill, with Jack next to Jill, Jack in front of Jill

which is 6!

or the number of permutations of these 6 "things":

Ann
Barbara
Carl
Doug
Edith
Jill&Jack, with Jill next to Jack, Jill in front of Jack

which is also 6!

Answer 7!-2×6! = 5040-2×720 = 5040-1440 = 3600 ways.

Edwin