Question 6: How many ways can you arrange 8 people at a round table (clockwise and anticlockwise arrangements

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Question 6: How many ways can you arrange 8 people at a round table (clockwise and anticlockwise arrangements are considered different)

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**Question 6: How many ways can you arrange 8 people at a round table (clockwise and anticlockwise arrangements are considered different)**

*   **a) With no restrictions**
    For circular permutations where clockwise and anticlockwise arrangements are considered different, the number of ways to arrange $n$ distinct items is $(n-1)!$.
    Here, $n=8$.
    Step 1: Apply the formula for circular permutations.
    $$(8-1)! = 7!$$
    Step 2: Calculate the factorial.
    $$7! = 7 	imes 6 	imes 5 	imes 4 	imes 3 	imes 2 	imes 1 = 5040$$
    There are $\boxed{	ext{5040 ways}}$ to arrange 8 people with no restrictions.

*   **b) If George and Harry must sit next to each other**
    Step 1: Treat George and Harry as a single unit. Now we are arranging $8-1=7$ units (6 individuals + the GH pair).
    Step 2: Arrange these 7 units around the table. The number of ways is $(7-1)! = 6!$.
    Step 3: Consider the internal arrangement of George and Harry within their unit. They can sit as (George, Harry) or (Harry, George), which is $2!$ ways.
    Step 4: Multiply the arrangements of the units by the internal arrangements of the pair.
    $$6! 	imes 2! = (720) 	imes (2) = 1440$$
    There are $\boxed{	ext{1440 ways}}$ if George and Harry must sit next to each other.

*   **c) If Mary refuses to sit next to George**
    Step 1: Calculate the total number of arrangements with no restrictions (from part a), which is $7! = 5040$.
    Step 2: Calculate the number of arrangements where Mary *does* 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 = $6! 	imes 2! = 720 	imes 2 = 1440$.
    Step 3: Subtract the arrangements where Mary sits next to George from the total arrangements.
    $$5040 - 1440 = 3600$$
    There are $\boxed{	ext{3600 ways}}$ 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, $n=8$.
    Step 1: Apply the formula for symmetrical circular permutations.
    $$\frac{(8-1)!}{2} = \frac{7!}{2}$$
    Step 2: Calculate the factorial and divide by 2.
    $$\frac{5040}{2} = 2520$$
    There are $\boxed{	ext{2520 ways}}$ if clockwise and anticlockwise arrangements are considered equivalent.

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