How many ways are there to place $7$ people to $10$ seats around a circular table so that Alex and Bob in these $7$ people do not sit together?

How many ways are there to place $7$ people to $10$ seats around a circular table so that two people Alex and Bob in these $7$ people do not sit together?

There are two cases for arranging them such that either Alex and Bob together or not. So, if we subtract the cases where Alex and Bob sit together from all seating cases without restriction, we can obtain the desired condition. To do this,

• Firstly find the all seating cases without restriction: Firstly, place one of them to one of $10$ seats by only $1$ ways because there is no way to distinguish $10$ seats around a circular table. After that, we have $6$ people to place in $9$ seats, but now we can distinguish which seat is which by using the position of the placed person. If so, there are $P(9,6)$ ways to do it.

• The cases where Alex and Bob sit together like a block: Let's place them anywhere when they are adjacent, then we have $8$ seats to place $5$ people, which we can do in $P(8,5)$ ways, but Alex and Bob can interchange in their block, so there are $2!P(8,5)$ ways.

Then, the answer is $$P(9,6) - 2!P(8,5)=60,480-13,440=47,040$$

Is my solution correct ?

Yes your work is correct. Another approach -

Alex takes one of the seats. That leaves $7$ seats for Bob to choose from (except adjacent seats to Alex). Rest $5$ of them can be seated in $5$ of the remaining $8$ seats.

So the answer is $ \displaystyle 7 \cdot {8 \choose 5} \cdot 5! = 47040$