How many ways nine people can sit on these chairs if at least one of the chairs in row two must remain unoccupied? 
Question: find how many ways nine people can sit on these chairs if at least one of the chairs in row two must remain unoccupied.
Attempt: $9!+(9!)(10C9)(6)+(9!)(11C9)(3)=82010880$
Reasoning: if three chairs are unoccupied, then there are $9!$ ways, if two chairs are unoccupied, there are $(9!)(10C9)(6)$ {the $6$ because the possible ways the unoccupied chairs are arranged}, $(9!)(11C9)(3)$ if one chair is $(9!)(11C9)(3)$
Correct answer: $59512320$
Could someone please help me understand what I am doing incorrectly?
 A: Alternatively, you can use complimentary method.
Number of total arrangements $ = \displaystyle {12 \choose 9} \cdot 9! = 79833600$
Number of arrangements in which all four seats of the second row are occupied $ = \displaystyle {9 \choose 4} \cdot 4! \cdot {8 \choose 5} \cdot 5! = 20321280$
(You choose four people out of nine and arrange them in second row. Then you arrange the remaining people in eight seats from the first and third rows)
So number of arrangements where at least one of the seats in the second row is not occupied
$$ = 79833600 - 20321280 = 59512320$$
A: I take it that there are four chairs in each "row"
You can apply inclusion-exclusion to avoid double counting viz
$C(4,1)*P(11,9) -C(4,2)*P(10,9) + C(4,3)*P(9,9) = 59512320$
Another way
P(row $2$ not full)$\times$(All permutations)
$=\left(1- \frac9{12}\frac 8{11}\frac7{10}\frac69 \right)\times P(12,9) =\frac{41}{55}\times P(12,9) = 59512320$
A: It is a matter of arranging the unoccupied chairs for each case. First, line up the $9$ people in any of the $9!$ ways.
If there is only one occupied chair in row $2$, then there are $4$ ways to position it.
If there are $2$ occupied chairs, then there is $1$ space in either row $1$ or row $3$, and $6$ ways in row $2$.
If there are $3$ occupied chairs, there are $\binom{8}{2}=28$ ways for the spaces in rows $1$ and $3$, and $4$ in row $2$.
So the final answer is $9!(4+8\cdot6+28\cdot4)=59,512,320$.
