How many ways can 12 boys be seated around a round table if two particular boys must sit together? 2. How many ways can six board of directors be seated on a circular table?

Here are the solutions to your assignment questions:

1. Seating 12 boys around a round table with 2 particular boys together

Step 1: Treat the 2 particular boys who must sit together as a single unit.
This reduces the number of entities to arrange to $(12 - 2) + 1 = 11$ units.

Step 2: Calculate the number of ways to arrange these 11 units around a circular table.
For circular permutations, the number of ways to arrange $n$ distinct items is $(n-1)!$.
So, for 11 units, the number of arrangements is $(11-1)! = 10!$.

$$10! = 3,628,800$$

Step 3: Account for the arrangements of the 2 particular boys within their unit.
The 2 boys can sit in $2!$ ways within their unit (Boy A then Boy B, or Boy B then Boy A).

$$2! = 2 \times 1 = 2$$

Step 4: Multiply the arrangements from Step 2 and Step 3 to get the total number of ways.
Total ways = (arrangements of units) $\times$ (arrangements within the unit)

$$3,628,800 \times 2 = 7,257,600$$

The 2 particular boys must sit next to each other in $\boxed{\text{7,257,600 ways}}$.

2. Seating six board of directors on a circular table

Step 1: Identify the number of people to be arranged.
There are $n=6$ directors.

Step 2: Apply the formula for circular permutations.
The number of ways to arrange $n$ distinct items in a circle is $(n-1)!$.

$$(6-1)! = 5!$$

Step 3: Calculate the factorial.

$$5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$$

Six board of directors can sit on a circular table in $\boxed{\text{120 ways}}$.

3. Expand the given expressions