SOLUTION: An algebra class has 20 students and 20 desks. For the sake of variety, students change the seating arrangement each day. How many days must pass before the class must repeat a sea

Question 1170760: An algebra class has 20 students and 20 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?
days must pass before a seating arrangement is repeated.
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?
There are seating arrangements that put them in the front seats.
What is the probability that Larry, Moe, Curly and Shemp are sitting in the front seats?
The probability is .

Answer by math_tutor2020(3817) (Show Source): You can put this solution on YOUR website!

Let
A = number of ways to arrange the students with no restrictions
B = number of ways to arrange the students with Larry, Moe, Curly, and Shemp in the front seats.

There are 20 seats and 20 students. This leads to 20! different ways to arrange them. The exclamation mark indicates factorial.

Recall that something like 7! means 7*6*5*4*3*2*1. We start with the given value, and count our way down, multiplying along the way until we reach 1. In the case of 20!, we'll start at 20, and count down to 1, multiplying along the way.
20! = 20*19*18*...*3*2*1
As you can probably guess, the result is very large.
We won't multiply these values out because you'll see later that a lot of the factors cancel.

So there are A = 20! different ways to arrange the students such that there are no restrictions. In other words, Larry, Moe, Curly, and Shemp may or may not be in the front seats.

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If we put them in the front row, there are 4! = 4*3*2*1 = 24 ways to arrange these four students. The remaining 20-4 = 16 students are assigned to the 16 remaining desks. There are 16! ways to arrange these remaining students.

So we have B = 4!*16! different ways to arrange the students such that Larry, Moe, Curly, and Shemp are in the front seats.
I won't multiply these values out either because we'll have a cancelation of a factor.

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In summary so far, we found

A = 20! ways to arrange the students (no restrictions)
B = 4!*16! ways to arrange the students such that Larry, Moe, Curly, and Shemp are in the front seats

To get the probability we want, we divide the two expressions
(A)/(B)
(4!*16!)/(20!)
(4!*16!)/(20*19*18*17*16!)
(4!)/(20*19*18*17) .... note the "16!"s cancel
(4*3*2*1)/(20*19*18*17)
(24)/(116,280)
(24*1)/(24*4845)
1/4845
0.00020639834881
0.000206

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Answer as a fraction = 1/4845
Answer as a decimal = 0.000206 (approximate)
Answer as a percentage = 0.0206% (approximate)

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