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?
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.
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...