Circular Arrangement with Two Specified Objects Not Together

On a circular table, if you pick any three people (D, C, and E), C will always be sitting between D and E regardless of how many empty seats separate them. This is because a circle has no endpoints.

Therefore, this condition is automatically satisfied in every single possible circular arrangement of the 6 people.

• Total unconstrained circular arrangements of 6 people = $$(6 - 1)! = 5! = 120$$ ways.

To find the number of ways where A and B sit next to each other, we tie them together into a single bundle: (AB).

• Now, we have 5 units to arrange around the circular table: (AB), C, D, E, and F.
• Circular arrangements of 5 units = $$(5 - 1)! = 4! = 24$$ ways.
• Inside their bundle, A and B can swap seats in $$2! = 2$$ ways (AB or BA).
• Total arrangements where A and B sit together = $$24 \times 2 = 48$$ ways.

To find the number of ways where A and B are never adjacent, we simply subtract the bad arrangements from the total arrangements:

$$\text{Valid Arrangements} = \text{Total Arrangements} - \text{Arrangements where A and B sit together}$$  $$\text{Valid Arrangements} = 120 - 48 = 72$$

Since 72 is the correct answer