12 guests at a dinner party are to be seated along a circular table

Question

Question: $12$ guests at a dinner party are to be seated along a circular table. Suppose that the master and mistress of the house have fixed seats opposite to one another. There are two specified guests who must always be placed next to one another. What is the number of ways in which the company can be placed?

My Method:
Let us first seat the $12$ guests. $2$ guests are always beside each other hence considering $11$ guests, the no. of combinations possible are $11!$. But the $2$ people can be arranged in $2$ ways, so total no. of combinations is given by $2*11!$.
Now the master can be seated between any $2$ people except the pair to be seated together. Hence, the no. of possibilities to seat him becomes $11$. The mistress is always opposite to him hence she would not contribute to the no. of total ways.

This gives,
The total no. of ways $=2*11*11!$

The Answer given:
First we seat the $2$ specified people in $2*10$ ways and the remaining $10$ people can be arranged in $10!$ ways. So total no. of ways $=2*10*10!$

But I don't understand what does the answer state and how my way is wrong. Can anybody help me understand this...

Answer 1

The book answer is correct

There is the master, mistress and $12$ guests, and in the absence of information to the contrary, it is customary to take chairs to be unnumbered, so there is only one way to seat the hosts opposite to each other.

In effect, the remaining $12$ seats become numbered with respect to the mistress at the $12$ o'clock position, say, and the master at $6$ o'clock position.

In between there are $6$ seats in each half, but the two specified people can only sit together in $2\cdot5$ ways in each half, i.e. $2\cdot10$ ways altogether.

The remaining $10$ guests can then be arranged in $10!$ ways

Putting everything together, ans $= 2*10*10!$

Answer 2

The mistress is always opposite the master, therefore not only can the master not sit between the two specified guests, but he cannot sit between the pair opposite them, either. Also, before someone sits down, the round table has no "first" or "last" position, so each of your $11!$ options has $11$ identical rotations. In general, the number of ways to arrange $n$ objects in a circle is $(n-1)!,$ not $n!,$ because all positions in which the first object can be placed are indistinguishable.

To make this clearer, consider the simple case of $3$ objects. There are of course $3!=6$ ways of arranging them in a row: $$1.abc$$$$2.acb$$$$3.bac$$$$4.bca$$$$5.cab$$$$6.cba$$ However, when we put the objects in a circle, options $1,4,$ and $5$ are the same, as are $2,3,$ and $6$, so there are only $2!=2$ different circular arrangements.