To solve this, we use the concept of complementary counting and consider the arrangement of the 4 friends in the available seats while ensuring at least one of them occupies an aisle seat.

Total Possible Arrangements:

• There are 11 seats available, and the 4 friends can choose any 4 out of these 11 seats. The number of ways to select 4 seats is given by the combination formula: $\binom{11}{4} = \frac{11 \cdot 10 \cdot 9 \cdot 8}{4 \cdot 3 \cdot 2 \cdot 1} = 330$
• Once the seats are chosen, the 4 friends can permute themselves in these seats in $4!$ ways: $4! = 24$
• Total arrangements: $330 \cdot 24 = 7920$

Complement: Arrangements with No One in an Aisle Seat

If no one sits in an aisle seat, the 4 friends must sit in the remaining $11 - 2 = 9$ non-aisle seats. The number of ways to select 4 seats from these 9 is: $\binom{9}{4} = \frac{9 \cdot 8 \cdot 7 \cdot 6}{4 \cdot 3 \cdot 2 \cdot 1} = 126$

• These 4 friends can permute themselves in $4!$ ways: $4! = 24$
• Total arrangements with no one in an aisle seat: $126 \cdot 24 = 3024$

Subtract Complement from Total:

The number of arrangements where at least one person sits in an aisle seat is: $7920 - 3024 = 4896$

Final Answer:

There are 4,896 ways for the 4 friends to arrange themselves such that at least one of them sits in an aisle seat.

Would you like a detailed breakdown or any clarification? 😊

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Related Questions:

1. How would the calculation change if there were more aisle seats available?
2. What if the number of friends increased to 5? How would the process change?
3. How does the concept of complementary counting help simplify problems like this?
4. How do permutations differ from combinations in problems like this?
5. Can this problem be solved using probability instead of counting? If so, how?

Tip:

Always consider the complement in counting problems—it’s often easier to calculate what you don’t want and subtract it from the total!