Answer:
40,320 ways.
Step-by-step explanation:
This problem involves permutations.
Analyze the problem: there are eight adjacent seats in the front row in order to be occupied by eight students.
Imagine that all eight students are still standing and have not had their seats yet. For the first seat, there are 8 possible choices on which student sits first. Let's name the first student S1.
Now that S1 is already seated, there are only 7 possible choices of who sits on the second seat. Let's name the second student S2.
Now that two students are already seated in the first and second seat, there are only 6 possible choices for the third seat. The student sitting on the third seat will be named S3.
This pattern continues until every student already has their seat.
F = { S1, S2, S3, S4, S5, S6, S7, S8 }
Now, introduce mathematics. Treat the group of eight students as a set, named set F above. The number of ways they can be arranged is n!
This can be obtained by calculating the number of possible choices per subsequent seat.
(choices for seat 1) × (choices for seat 2) × ... × (choices for seat 8).
This simplifies to be 8! or 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1.
The product will be 40,320.
Therefore, there will be a total of 40,320 ways to arrange eight students among eight seats.
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CTTO-
