Question 1122549
The number of all possibilities: 7P7=7!


The number of the cases where Jill has the first chair: 6P6(since Jill has had the first chair, the variable is just the rest 6 chairs)=6!
P(Jill in first chair)=6!/7!=1/7.


The number of the cases where Jill has the first and Jack has the second: 5P5=5!
P(Jill in first and Jack not in second)=(6!-5!)/7!=5/42.


And we're done.