A question in permutation Help is needed in solving the following problem.
$8$ persons ($A$ and $B$ and $P, Q, R, S, T, U$) are to be seated in $2$ rows ($4$ seats per row). Find the number of ways that $A$ and $B$ are sitting in the same row.
My way of doing it is:-


*

*Tie $A$ and $B$ together as one object called $(A+B)$.

*Suppose that $(A+B)$ sits on seat-$1$ and seat-$2$ like $(A+B)$ _ _ | _ _ _ _. Then there are $6!$ ways of allocating the remaining $6$ persons to sit on the remaining seats

*Of course, it is possible to have arrangements like _ $(A+B)$ _ | _ _ _ _. There are a total of $6$ such arrangements including the one mentioned in step 2.

*Now, if we untie $(A+B)$ so that they can sit freely in the same row. It is a permutation of $4$ taken from $4$ giving a total of $4!$.

*All of the above sum up to $6! × 6 × 4!$
This is obviously wrong because $6! × 6 × 4! \gt 8!$, the total number of possible number of arrangements.
I know the place ran wrong is allowing $(A+B)PQ$ and $P(A+B)Q$ and $PQ(A+B)$ being counted more than once.
But I don’t know how to fix it. This is my first question.
Should this problem be solved in some other ways?
 A: There are 8 choices for A's seat, and then 3 choices for where B can sit if B is in the same row as A.
The remaining 6 people can be seated in $6!$ ways, so we have $8\cdot3\cdot6!=17,280$ possibilities.
A: I would go about doing this by counting the complement, i.e. $A$ and $B$ are sitting in different rows. Without loss of generality, assume that $A$ is sitting in the first row. Then, we pick three people out of the six to six with $A$ and permute. We also permute the four people in the second row. Hence, there are
$$\binom{6}{3} \times 4! \times 4!$$
ways of doing this. But, then we could have picked $B$ to sit in the first row, so there are a total of
$$2 \times \binom{6}{3} \times 4! \times 4!=23040$$
ways of seating where $A$ and $B$ are sitting in different rows. Hence, there are
$$8! - 23040 = 17280$$
ways of seating where $A$ and $B$ are sitting in same rows.
Or, we can do this directly also. Pick a row that both $A$ and $B$ are going to sit in (two choices), then pick the two people that they are going to sit with from six. Permute those four, and permute the other four in the other row. Hence, there are
$$2 \times \binom{6}{2} \times 4! \times 4! = 17280$$
ways!
(It turns out my way was longer...)
A: We count the ways to place the people with A and B not in the same row.  There are $2$ ways to decide which row A is in, then $4 \times 4$ ways to place A and B, then $6!$ ways to place the remaining people.  
So the number of ways with A and B in the same row is
$$8! - 2 \times 4 \times 4 \times 6! = 17,280$$
