1) In how many ways can 12 distinct objects be separated into three equal piles?

2) In how many ways can 25 students be assigned to 4 distinguishable study groups if at least 6 students must be in each group?

3) in how many ways can 4 students be assigned to 9 class periods if:
a) there are no restrictions?
b) each students must go to a different period?

1.

There are n=12! ways to line up 12 distinct object into 12 ordered positions.

If positions 1,2,3 form one pile, then N would be overcounted by 3! times. Therefore n must be divided by 3! for each group of 3.

The total number of ways is therefore
12!/(3!3!3!3!)

2. Follow the same argument as in Q1 to place 24 (distinct) students into 4 groups of 6. There are 4 ways to place the 25th student, so multiply by 4.

3.
No restriction:
We give the choice to the students.
Number of choices for the first student=9
number of choices for the second student=9
.....
number of choices for the fourth student=9
Use the multiplication rule to establish the total number of ways.

3.
With restriction that they go to a different session:
Number of choices for the first student=9
Number of choices for the second student = 8
.....
number of choices for the 4th student = 6
Use the multiplication principle to get the total number of arrangements.

1) To separate 12 distinct objects into three equal piles, we can use the concept of combinations. Each object can be assigned to one of the three piles.

Since there are three piles and each object can be placed in one of them, the total number of ways to separate the objects can be calculated as 3^12, which gives us 531,441 ways.

2) In this scenario, we have 25 students who need to be assigned to 4 distinguishable study groups, with each group needing to have at least 6 students.

We can solve this using a combination formula. We'll start by assuming that we have 6 students in each group. This leaves us with 25 - (4 * 6) = 1 student remaining. We'll temporarily assign this student to any one of the four groups.

Now, we need to divide the remaining 1 student among the groups. We can choose any of the four groups to accommodate the remaining student, so there are 4 possible choices.

Therefore, the total number of ways to assign 25 students to 4 groups while maintaining at least 6 students in each group is 4.

3a) If there are no restrictions on the assignment of 4 students to 9 class periods, we can solve this using the concept of permutations. Each student can be assigned to any one of the 9 class periods.

Therefore, the total number of ways to assign the 4 students to the 9 class periods is 9^4, which is 6561.

3b) If each student must go to a different period, we can solve this using a permutation formula. For the first student, we have 9 class periods to choose from. For the second student, we have 8 remaining class periods to choose from. Similarly, for the third student, we have 7 remaining choices, and for the fourth student, we have 6 choices.

Therefore, the total number of ways to assign the 4 students to the 9 class periods, making sure each student goes to a different period, is 9 * 8 * 7 * 6 = 3024.

1) To find the number of ways to separate 12 distinct objects into three equal piles, we can use the concept of combinations. Each pile should contain 12/3 = 4 objects.

To solve this, we can follow these steps:
- First, choose 4 objects from the 12 to put in the first pile. This can be done in C(12, 4) ways.
- Then, choose 4 objects from the remaining 8 for the second pile. This can be done in C(8, 4) ways.
- Finally, there is only one combination left for the third pile, as whatever is left after the first two choices will automatically go into the third pile.

So, the total number of ways to separate 12 distinct objects into three equal piles is given by:
C(12, 4) * C(8, 4) = (12! / (4! * (12-4)!)) * (8! / (4! * (8-4)!)) = 34650 ways.

2) To find the number of ways to assign 25 students to 4 distinguishable study groups, where each group must have at least 6 students, we can use combinations along with the concept of partitions.

To solve this problem, we can follow these steps:
- First, distribute 6 students to each group. This can be done in C(25, 6) * C(19, 6) * C(13, 6) ways.
- After distributing 6 students to each group, there will be 25 - (6 * 4) = 1 student remaining.
- We can distribute this last remaining student to any of the 4 groups. So, there are 4 choices for the last student.

Therefore, the total number of ways to assign 25 students to 4 groups with at least 6 students in each group is given by:
C(25, 6) * C(19, 6) * C(13, 6) * 4 = (25! / (6! * (25-6)!)) * (19! / (6! * (19-6)!)) * (13! / (6! * (13-6)!)) * 4 = 2,197,736 ways.

3) a) If there are no restrictions on how 4 students can be assigned to 9 class periods, this can be approached using the concept of permutations.

To solve this, we need to find the number of ways to arrange 4 students in 9 different class periods. Each student can be assigned to any class period, and multiple students can be assigned to the same class period.

Therefore, the total number of ways to assign 4 students to 9 class periods without any restrictions is given by:
9^4 = 6561 ways.

b) If each student must go to a different period, this can be approached using permutations again, but with constraints.

To solve this, we need to find the number of ways to arrange 4 students in 9 different class periods such that no two students are assigned to the same class period.

Therefore, the total number of ways to assign 4 students to 9 class periods without repetition is given by:
P(9,4) = 9! / (9-4)! = 9! / 5! = 9 * 8 * 7 * 6 = 3024 ways.