6. The vertices of an octagon will be labeled with the letters A. B. C. D. E. F. G. and H. not necessarily in that order. If one vertex is labeled A, which of the following is equal to the number of possible orderings of the labels for the remaining vertices?
Answer: A
The number of possible orderings of the labels for the remaining vertices is 7!
When one vertex of the octagon is labeled A, there are 7 remaining vertices (B, C, D, E, F, G, and H) that need to be labeled. The number of possible arrangements for these 7 vertices is given by 7 factorial (7!).
A) 7!
This option is correct because once one vertex is designated as A, there are exactly 7 remaining vertices that can be arranged in any order, which is calculated as 7!.
B) 8!
This option is incorrect as it represents the total arrangements of all 8 vertices without fixing any label. Since one vertex is already labeled A, calculating arrangements for all 8 vertices is not applicable.
C) 7!/8
This option is incorrect because it suggests a division of the arrangements by 8, which is not relevant to the problem. The total arrangements should simply account for the 7 remaining vertices.
D) 8!/7
This option is incorrect as it implies a calculation for the arrangements of all vertices divided by 7, which does not apply here. We only need the arrangements of the 7 vertices left after fixing A.
E) 8!/7!
This option is incorrect because it simplifies to 8, representing the number of ways to choose one vertex from 8, rather than the arrangements of the remaining 7 vertices.
Conclusion
The correct answer, 7!, accurately reflects the number of ways to arrange the remaining vertices after one vertex has been fixed as A. All other options fail to account for the specified condition and lead to incorrect calculations based on the initial configuration of the octagon.