Arrangement of $12$ people in a row such that neither of $2$ particular persons sit on either of $2$ ends of the row 
If $12$ persons are arranged in a row such that neither of two particular persons can sit on either end of the row, is

My attempt:
Total ways $=$ Sitting $12$ persons in a row $-$ Sitting $2$ particular persons $2$ ends of the row
$$=12!-2!\cdot 10!$$
But it seems that this answer is wrong.
Can anyone please explain to me the right answer. Thanks.
 A: You have only removed from your count where both of your people sit at the ends and have neglected to remove where only one of your people sit at the ends but not the other.
Rather than thinking in terms of the perspective of the seats... we can think with respect to the people instead and approach with rule of product like usual.

*

*Pick which seat the younger of the two special people sit (noting it cannot be either end):  10 choices


*Pick which seat the older of the two special people sit (noting it cannot be either end or where the first person sat): 9 choices


*Pick which seat the youngest of the remaining people sit (noting it cannot be where either of the first two people sat but now may include the ends): 10 choices


*Pick which seat the next youngest of the remaining people sit (similar restrictions): 9 choices


*Continue in this fashion seating each of the rest of the people in sequence
You will find then that there are $10\cdot 9\cdot 10!$ total arrangements.
A: Excluding the seats at each end, there are ten seats that these two people can be placed in, with $T_{9}= \frac{9(9+1)}{2}=45$ choices for this.
They can do this in either order, so we double the result.
The remaining ten seats can be filled in $10!$ ways with the other ten people.
This gives a total of $10!×45×2=10!×90$ arrangements meeting the conditions you have.
A: What the OP subtracted from $12!$ is the number of arrangements in which both ends are occupied by the two particular people, but the problem wants us to eliminate the arrangements in which either end is occupied by one of them. This is more easily counted directly, by first seating the two particular people in the interior $10$ seats, which can be done in $10\cdot9$ ways, then seating the remaining $10$ people willy nilly, which can be done in $10!$ ways, for a total of
$$10\cdot9\cdot10!$$
arrangements.
A: $12! - (2! . 10!)$ is the number of ways $12$ people can be arranged, excluding the cases where person A and person B are both at the ends at the same time.
You haven’t excluded the cases where person $A$ is at one end and person $B$ isn’t at the other end, and same with $A <-> B$
A: The $10$ normal persons can be seated in $10!$ ways. There are $9$ slots in between them for the  special persons. When the first special person is seated there are $10$ slots for the second special person. Therefore there are
$$10!\cdot  9\cdot 10=326\,592\,000$$
admissible seatings for the $12$ persons.
