Sorry, I forgot to add that I only need help with part II

We have two seats being tied with a red ribbon.

This makes the seats no longer "indistinguishable" in a rotary sense, since we can no longer rotate the people there to get an exact identical arrangement. Now, the question is no different from placing the people in a straight line.

So, we
- pick any seven people (10C7)
- then we arrange the 7 people (7!)

So, the total number of ways
= 10C7 times 7!
= 604800

Another way to see this is to first arrange the 10 people in an unrestricted fashion (10!)

But now, we remove three people. The three resultant empty seats are now considered identical (previously not identical since the three occupants were obviously distinguishable), so therefore there are repetitions by a factor of 3!.

To explain this, I give you one example as follows, I removed H, I and J from the list of 10 people occupying all 10 seats and place "x" instead to represent the newly vacated seats.

(Before removal of people)
ABCDEFGHIJ
ABCDEFGHJI
ABCDEFGIHJ
ABCDEFGIJH
ABCDEFGJHI
ABCDEFGJIH

(After removal of people)
ABCDEFGxxx
ABCDEFGxxx
ABCDEFGxxx
ABCDEFGxxx
ABCDEFGxxx
ABCDEFGxxx

But these are identical cases, and it's where the 3! comes from.