Question

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

• A. $20\cdot 10!$
• B. $22\cdot 10!$
• C. $44\cdot 10!$
• D. None

Answer 1

The correct option is A $20\cdot 10!$

Those two person can be arranged in the row in 5 ways and amongst themselves they can inter change

no. of ways 2 persons can be selected

2! = 2 people inter change

2 = there are 2 rows

10! = semantically 10 people can be sated on 10! ways

$\therefore$ no. of ways of arranging $=5\times 2!\times 2\times 10!$

$\therefore$ total no. of ways $=5\times 2!\times 2\times 10!$

$=20\times 10!$