permutation question (1 Viewer)

[Masaken]

How many ways can ten people be arranged in a line if neither of two particular people can sit on either end of the row?

Hey yall, I'm pretty stumped with this question. I know it's probably going to be easier if it gets explained in a much better way than the worked solutions of my textbook lol, but is it alright if you can lend a helping hand (or at least where to start? I'm a little confused on where to go from here lol).

 

[jimmysmith560]

Would the following working help?

Without any restrictions, there are 10! ways in which you can arrange them.

There are 9! × 2 ways where the first person is at any end 9! × 2 ways where the second person is at any end.

There are 8! × 2 ways where both persons are at end

You can subsequently use the inclusion-exclusion to get the answer:

10! − (9! × 2 + 9! × 2) + 8! × 2 = 2257920, matching the answer in the textbook.

 

[Masaken]

There are 9! × 2 ways where the first person is at any end 9! × 2 ways where the second person is at any end.

omg i forgot about this method

also, regarding the quoted, would this situation be where only one person is at any of the ends at any one time?

 

[jimmysmith560]

omg i forgot about this method

also, regarding the quoted, would this situation be where only one person is at any of the ends at any one time?

Yes I believe that is correct in this particular case.

 

[Masaken]

Yes I believe that is correct in this particular case.

okay thank you! that makes a lot more sense