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.

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.
