How many ways can you arrange 8 people at a round table if clockwise and anticlockwise arrangements are considered different? Consider cases with no restrictions, with George and Harry sitting together, with Mary refusing to sit next to George, and with clockwise/anticlockwise arrangements considered equivalent.

$$
7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 5040
$$
There are $\boxed{\text{5040 ways}}$ to arrange 8 people with no restrictions.

• b) If George and Harry must sit next to each other

Now we are arranging $8-1=7$ units (6 individuals + the GH pair).

The number of ways is $(7-1)! = 6!$.

They can sit as (George, Harry) or (Harry, George), which is $2!$ ways.

$$
6! \times 2! = (720) \times (2) = 1440
$$
There are $\boxed{\text{1440 ways}}$ if George and Harry must sit next to each other.

• c) If Mary refuses to sit next to George

This is similar to part b.
Treat Mary and George as a single unit. We are arranging $8-1=7$ units. This gives $(7-1)! = 6!$ ways.
Within the unit, Mary and George can be arranged in $2!$ ways (MG or GM).
So, arrangements where Mary does sit next to George = $6! \times 2! = 720 \times 2 = 1440$.

$$
5040 - 1440 = 3600
$$
There are $\boxed{\text{3600 ways}}$ if Mary refuses to sit next to George.

• d) If clockwise and anticlockwise arrangements are considered equivalent. For circular permutations where clockwise and anticlockwise arrangements are considered equivalent (symmetrical arrangements), the number of ways to arrange $n$ distinct items is $\frac{(n-1)!}{2}$. Here, $n=8$.