Question

There are $12$ guests at a dinner party. Supposing that the master and mistress of the house have fixed seats opposite to each other and that there are two specific guests who must always be placed next to one another: the number of ways in which the company can be placed is

• A. $20.10!$
• B. $22.10!$
• C. $44.10!$
• D. $none$

Answer 1

Answer: (A)

$20.10!$

Given in question we have master, mistress and $12$ guests. So, let us put master and mistress opposite to each other. Now, in between them $6$ seats are available in each half but the two specification people can seat in $2.5$ ways in each half so total of $2\times 2\times 5$ ways.

Now we have remaining $10$ guests so the total no. of ways:

$=2\times 2\times 5\times 10!$

$=20.10!$

Hence, the answer is $20.10!.$