Question

An algebra class has 8 students and 8 desks. For the sake of variety, students change the seating arrangement each day. How many days must pass before the class must repeat a seating arrangement?

Suppose the desks are arranged in rows of 4. How many seating arrangements are there that put Larry, Moe, Curly, and Shemp in the front seats ?

What is the probability that Larry, Moe, Curly and Shemp are sitting in the front seats? ​

Answer 1

The algebra class with 8 students and 8 desks has 40,320 different seating arrangements, so they can switch seats for 40,320 days without repeating. With four specific students in the front, there are 24 possible arrangements for them, with a probability of approximately 0.05952% for this scenario.

Step-by-step explanation:

The algebra class seating arrangement is a permutation problem where each of the 8 students can sit in any of the 8 desks. The number of unique seating arrangements possible is determined by 8 factorial (8!), which is 40,320 different seating arrangements. Thus, it would take 40,320 days to cover all possible seating arrangements without repetition.

For the second part of the question, when considering Larry, Moe, Curly, and Shemp occupying the four front seats, this arrangement is another permutation problem. There are 4 factorial (4!) ways to arrange these four students in the front seats, resulting in 24 possible arrangements.

To find the probabilitythat these four specific students will sit in the front seats on any given day, we divide the number of favorable arrangements (24) by the total number of seating arrangements (40,320), resulting in a probability of 24/40,320, which simplifies to 1/1,680, or approximately 0.0005952, which can be expressed as 0.05952%.