Arranging people in a line without specific people at either end

Question

In 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?

What i thought was find how many ways one particular person must sit at either end then multiply that value by 2 then subtract it from how many ways ten people can be arranged without restriction

Answer 1

In 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?

Without any restrictions, $10!$ ways you can arrange them

$9!\times 2 $ ways where first person is at any end $9!\times 2 $ ways where second person is at any end

$8!\times 2 $ ways where both persons are at end

Then use inclusion-exclusion for answer
$10! - (9!\times 2 + 9!\times 2) + 8!\times 2 $

Answer 2

You have $8$ choices for the person sitting on one end (because there are only $8$ people that can be on either end), you have $7$ choices for the person sitting on the other end (because you already put a person on the end and you still can't put two people on the end), and then you have $8!$ for the rest of the people. Your total is therefore $8*7*8!$.