A number of n students attend the lecture of probabilities

[dionys]

Hi guys.
=====================================================================
Can you help me with these problems in combinations.I don't know if my solutions are correct.
=====================================================================

Problem 1
=========
There are 3 classes,each one with n students.We must select 3 students from the total of 3n students.
i)How many are the possible choices?
C=combinations
The possible choices are:
C(n,3)*C(n,0)*C(n,0)*3 + C(n,2)*C(n,1)*C(n,0)*3! + C(n,1)*C(n,1)*C(n,1)

ii)How many are the possible choices,if the 3 students belong to the same class?
C(n,3)*C(n,0)*C(n,0)*3

iii)How many are the possible choices,if 2 of the 3 students belong to the same class and the third to a different class?
C(n,2)*C(n,1)*C(n,0)*3!

iv)How many are the possible choices,if the 3 students belong to a different class?
C(n,1)*C(n,1)*C(n,1)

v)Use your answers i)-iv) to expand C(3n,n)
I don't know this one

=====================================================================

Problem 2
=========
A number of n students attend the lecture of probabilities.The array with the results of the final exam includes only the names of the students that passed the lesson in descending order.We assume that there arent any students with the same grade.How many are the possible arrays?

C(n,1)+C(n,2)*2!+C(n,3)*3!+...+C(n,n)n!

=====================================================================

Problem 3
=========
We put the hats of n persons in a box.Then each person randomly chooses a hat from the box.Find the probability that n-2 persons
will accurately choose their own hat.

I guessed that n-2 persons choose acurately their own hat is the same
with: 1 - 2 persons don't take their own hat
=[1 - C(n,1)*(n-1)!+C(n-1,1)*(n-2)!] / n!
=[1-n!+(n-1)!]n!?i'm not sure?
=====================================================================

 

[member 553083]

For the first problem, you need to use the formula C(n,r) = n! / (r! * (n-r)!). This is the formula for calculating the number of combinations. For the second and third problems, you can use the same formula but with different parameters. For example, in the second problem, you can use the formula C(n,r) = n! / (r! * (n-r)!), where n is the number of students and r is the number of students that passed the exam. For the third problem, you can use the formula C(n,r) = n! / (r! * (n-r)!), where n is the number of people and r is the number of people that don't take their own hat. Hope this helps!

 

[wais]

Hi there,

First, let me say that it's great that you are actively working on these problems and seeking help if needed. It shows determination and a desire to improve. I'll do my best to provide some guidance and corrections to your solutions.

Problem 1:
i) Your solution for the first part is correct. The possible choices are C(n,3) for selecting 3 students from one class, multiplied by C(n,0) for selecting 0 students from the other two classes, multiplied by 3 for the number of ways to choose which class the 3 students come from. However, your solution for the second part is incorrect. Since the 3 students must belong to the same class, the second and third terms should be C(n,0) instead of C(n,1). So the correct solution should be C(n,3)*C(n,0)*C(n,0)*3.

ii) Your solution for this part is correct. Since 2 of the 3 students must belong to the same class, we have C(n,2) for selecting 2 students from one class, multiplied by C(n,1) for selecting 1 student from another class, multiplied by 3! for the number of ways to arrange the 3 students.

iii) Your solution for this part is also correct. We have C(n,1) for selecting 1 student from one class, multiplied by C(n,1) for selecting 1 student from another class, multiplied by C(n,1) for selecting 1 student from the third class.

iv) Your solution for this part is correct as well. We have C(n,1) for selecting 1 student from each of the 3 classes.

v) To expand C(3n,n), we can use the binomial theorem. So C(3n,n) = (3n)! / (n!)^2.

Problem 2:
Your solution for this problem is incorrect. The correct solution should be n!. This is because for the first student, there are n possible choices, for the second student there are n-1 possible choices, and so on. So the total number of possible arrays is n * (n-1) * (n-2) * ... * 2 * 1 = n!.

Problem 3:
Your solution for this problem is also incorrect. The correct solution should be (n-2)! / n!. This is because