How many ways are there to seat $10$ people consisting of $5$ couples in a row of 10 seats, if all couple are to get adjacent seats.
I proceeded by creating five- 2 block partitions of the $10$ seats by $\frac{10!}{2!^5*5!}$ and then distributing these partitions of $2$ seats each to $5$ couples in 5! ways accounting to $\frac{10!}{2!^5}=113400$ but it came to be incorrect.
Where I went wrong?
First fill $10$ slots (seats) left to right keeping in mind the restriction that couples have to sit together. There are $10$ choices for the first slot (no restriction),$ 1$ choice for the second (since the first person’s mate has to be seated there), $8$ choices for the third slot (since $2$ people have now been seated, leaving $8$), $1$ choice for the fourth slot (since this seat has to go to the mate #$3$)
So, there are $10\times8\times6\times4\times2=3840$ ways