A group of 90 students is to be split at random into 3 classes of equal size. All partitions are equally likely. Joe and Jane are members of the 90-student group. Find the probability that Joe and Jane end up in the same class.

My idea:

the total combination: 90C30
since 2 have to be together then I only have 88 students to arrange in groups fo 30-30-28, and It'll be:

88!/(30! 30! 28!)

But then, should I multiply by 3 or how should I add the two that must remain together.

P = [(??) * 88!/(30! 30! 28!)] / 90C30

Or maybe I'm doing this wrong.

Method 1:

Assign a different number to each student ranging from 1 to 90, numbers 1 to 30 go in group 1, 31 to 60 go to group 2 61 to 90 go to group 3.

All possible partitions are obtained with equal probablity by a random assignment if these numbers, it doesn't matter with which students we start, so we are free to start by assinging a random number to Joe and then we assign a random number to Jane. After Joe has been assigned a random mumber there are 89 random numbers available for Jane and 29 of these will land her in the same group as Joe. Therefore the probability is 29/89.

Method 2:

We can consider the total number of ways of partitioning N objects into groups of n1, n2,...nr groups. There are

N!/(n1! n2!...nr!) ways to do that. But when two groups become of equal size we need to consider if we consider a permutation that has the effect of interchanging these groups as a different partition or the same partition. If we don't count that as diferent, then we need to divide by additional symmetry factors.

If we choose to do this, then the number of ways to make a partitioning where Joe and Jane end up in the same group is:

88!/(30!^2 2! 28! )

The total number of partitions is:

90!/(30!^3 3!)

Dividing the two numbers gives:

30!^3 3! 88!/(30!^2 2! 28! 90!) =

29/89

Method 3

The same as above, but now without symmetry factors for permuting groups. This amounts to labeling the 3 groups and it then matters in which group Joe and Jane end up in as these are counted as distinct possibilities.


The number of ways to make a partitioning where Joe and Jane end up in the same group is now:

3 88!/(30!^2 28!)

The total number of partitionings is now 90!/30!^3. Dividing the two numbers gives:

3 30!^3 88!/(30!^2 28! 90!) =

29/89

Method 4:

Similar as method 1 we just count all ways of placing the students in the 3 groups without taking into account that permuting the students in one group leads to the same partitioning.

There are then 3 ways to choose in which group Joe and Jane will go. And given that choice there are 30*29 ways to assing them a position in that chosen group. The total number of ways to divide all the students with Joe and Jane in the same group is thus:

3*30*29*88!

The total number of ways of partitioning the students is 90!, the ratio is:

3*30*29*88!/90! =

29/89

Well, it seems like you're doing a great job with the combinations so far! Let's see if we can figure out the rest.

First, let's find the total number of ways to split the 90 students into 3 equal-sized classes. To do this, we can calculate 90C30, which represents choosing 30 students out of 90 at a time. So, the total number of ways is 90C30.

Now, let's focus on the case where Joe and Jane are in the same class. We need to choose 28 more students for their class out of the remaining 88 students. The number of ways to do this is 88C28.

However, there are also 2 remaining classes to consider. After selecting the 30 students for Joe and Jane's class, we have 60 students left. So, we need to distribute these remaining 60 students into 2 classes of 30 students each. The number of ways to do this is 60C30.

Now, to calculate the probability, we divide the number of favorable outcomes (which is 88C28 * 60C30) by the total number of outcomes (which is 90C30).

P = (88C28 * 60C30) / 90C30

Hope that helps! Keep up the good work and let me know if you have any more questions.

To find the probability that Joe and Jane end up in the same class, we can approach it as follows:

1. Let's consider the total number of ways to divide the 90 students into 3 equal-sized classes. We can find this by calculating C(90, 30) (choosing 30 students for the first class), multiplied by C(60, 30) (choosing 30 students for the second class), and this will automatically determine the last class's composition.

2. Now, to find the favorable outcomes where Joe and Jane are in the same class, we need to fix them in one of the classes. We can consider this as fixing their position in the first class and dividing the remaining 88 students into 29 and 29 for the other two classes.

3. To calculate C(88, 29) * C(58, 29), we choose 29 students for the second class and 29 students for the third class from the remaining 88 students, while Joe and Jane are already accounted for in the first class.

4. Finally, the probability will be the favorable outcomes divided by the total outcomes.

So, the probability can be calculated as: (C(88, 29) * C(58, 29)) / (C(90, 30) * C(60, 30)).

Simplifying this expression will give you the probability that Joe and Jane end up in the same class.

To find the probability that Joe and Jane end up in the same class, you need to consider the different ways in which they can be placed in a class together and divide it by the total number of possible arrangements.

First, let's calculate the total number of ways to split the 90 students into 3 classes of equal size. Since all partitions are equally likely, we can calculate this using combinations. The total combination will be 90C30, which represents choosing 30 students for the first class, 30 students for the second class, and the remaining 30 students for the third class.

Now, let's consider the number of ways in which Joe and Jane can end up in the same class. We can treat Joe and Jane as a single entity since they have to be together, and consider them as one student. So, we now have 89 students to arrange in 3 classes: 88 regular students and Joe-Jane as one student. We can calculate this using combinations as well, choosing 29 students for the first class, 29 students for the second class, and the remaining 30 students for the third class.

The number of ways that Joe and Jane end up in the same class is: (88C29) * (58C29)

To find the probability, divide this number by the total number of ways to split the 90 students into 3 classes: (88C29) * (58C29) / (90C30).

So, the probability that Joe and Jane end up in the same class is: (88C29) * (58C29) / (90C30)