$12$ persons are to be seated at a square table, three on each side. $2$ persons wish to sit on the north side and two wish to sit on the east side. One other person insists on occupying the middle seat (which may be on any side). Find the number of ways they can be seated.
My attempt is as follows:-
Let's separate the $5$ persons with special demands.
Case $1$: Offer the middle seat of the north side to the person who want to sit on any of the middle seat.
Now offer the two seats of the north side to the two people who want to sit on the north side, they can arrange themselves in $2!$ ways
Offer the two seats of the east side to the two people who want to sit on the east side and choose $1$ person from remaining people and then arrange them $\implies \displaystyle{7\choose 1}3!$
Now arrange remaining 6 people in $6!$ ways. So total arrangements will be $2!\cdot3!\cdot6!\cdot\displaystyle{7\choose 1}$
Case $2$: Offer the middle seat of the east side to the person who want to sit on any of the middle seat.
Now all the calculations will remain same as it was in Case $1$ $\implies 2!\cdot3!\cdot6!\cdot\displaystyle{7\choose 1}$
Case $3$: Offer the middle seat of the south side or west side to the person who want to sit on any of the middle seat.
Offer the two seats of the north side to the two people who want to sit on the north side and choose $1$ person from remaining people and then arrange them $\implies \displaystyle{7\choose 1}3!$
Offer the two seats of the east side to the two people who want to sit on the east side and choose $1$ person from remaining people and then arrange them $\implies \displaystyle{6\choose 1}3!$
Then arrange remaining $5$ people in $5!$ ways. So total arrangement will be $\displaystyle{2\choose 1}\cdot3!\cdot3!\cdot{7\choose 1}\cdot{6\choose 1}\cdot5!$
Adding them up will give $2!\cdot3!(7!+7!+6\cdot7!)=2!\cdot3!\cdot8!$
Am I missing any short way here because in the last all calculations took the form of a very compact expression.