Four students fall asleep during class and, while asleep, the professor rearranges their notebooks so that when they wake up, nobody has his or her own notebook. In how many ways could the professor have rearranged the notebooks?
4
6
9
24
Explain
The correct answer is nine.
Student A can get any of three books: b, c or d. Assume he gets b. Then there are only three possibilities of distributing the remaining three (a, c and d) among students B, C and D. B can get a, c or d, but there is only one way to distribute the other two. The same logic applies to each of the three possible assignments to student A.
3 x 3 = 9
hey how come u didn't mind giving the correct answer this time drwls????
The hardest part of the problem for me was figuring out a method of solution. My first attempt, using factorial numbers, was wrong. So I took more time to think about it, and even made a list of possible rearragements. Computing the number was easy, once the logic was done properly.
Students who are only looking for a number to plug into a computerized test sheet are missing the fun of mathematics, and will probably never master the subject.
Let the students be abcd
and the exams ABCD
Here are the possible ways to distribute exams:
BCDA
BDAC
BADC
CADC
CDAB
CDBA
DABC
DCAB
DCBA
To find the number of ways the professor could have rearranged the notebooks, we can use the concept of permutations.
In this scenario, there are 4 students and 4 notebooks. As no student can have their own notebook when they wake up, we need to find the number of ways to assign the notebooks to the students so that no one gets their own notebook back.
To solve this problem, we can use the principle of derangements. A derangement is a permutation of a set where none of the elements appear in their original position.
The formula to calculate the number of derangements of n elements is given by the derangement formula: D(n) = n!(1 - 1/1! + 1/2! - 1/3! + ... + (-1)^n/n!).
For n = 4, the formula becomes D(4) = 4!(1 - 1/1! + 1/2! - 1/3! + 1/4!).
Now, let's calculate it step by step:
1. Calculate 4! (4 factorial): 4! = 4 x 3 x 2 x 1 = 24.
2. Calculate 1/1!: 1/1! = 1.
3. Calculate 1/2!: 1/2! = 0.5.
4. Calculate 1/3!: 1/3! = 1/6 = 0.1667.
5. Calculate 1/4!: 1/4! = 1/24 = 0.0417.
Now, plug these values into the formula:
D(4) = 4!(1 - 1 + 0.5 - 0.1667 + 0.0417)
= 24(1 - 1 + 0.5 - 0.1667 + 0.0417)
= 24(1 + 0.5 - 0.1667 + 0.0417)
= 24(1.375)
= 33.
Therefore, the number of ways the professor could have rearranged the notebooks is 33.
So, none of the given options (4, 6, 9, and 24) represents the correct answer.