# Arrangement of guests in dinner party

12 guests at a dinner party are to be seated along a circular table. 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) 20.10!

B)22.10!

C)44.10!

D)NONE

How I tried:

as the master and mistress are to be seated opposite to each other there are 2 possible ways (interchanging their position) of arrangement, now the two guests to be seated next to each other also have two possible ways (interchanging their position) now these two guests be treated as one guest and then arranging the 12-2-2+1=9 guests in remaining position= 9!

Therefore the total number of arrangement =2.2.9!=36.9!

But the answer is A)20.10!, where am I wrong, how to solve the problem?

. means multiply

• Does . mean multiply? Feb 8 '20 at 7:16
• Welcome to MSE. Please use MathJax to format the math in your posts. For example $2\cdot2\cdot9!$ is typeset as $2\cdot2\cdot9!$. Feb 8 '20 at 7:23

## 1 Answer

Seat the master. It does not matter where since two seating arrangements that differ by a rotation are considered to be equivalent. The mistress must sit opposite the master. Since there are $$12$$ guests, there are six seats between the master and mistress on each side. If we proceed clockwise around the table from the master, the block of two seats for the two guests who must sit together must either begin in one of the first five seats before we reach the mistress or in one of the first five seats after we reach the mistress. Choose one of those $$10$$ places for the block of two seats to begin. We have two ways to arrange the specified guests in the block of two seats. The remaining ten guests can be arranged in $$10!$$ ways in the remaining ten seats as we proceed clockwise around the table from the master. Hence, there are $$2 \cdot 10 \cdot 10!$$ admissible seating arrangements.