91Ó°ÊÓ

To distribute 8 teachers among 4 schools without any restrictions, we will use the combination formula: \[C(n, r) = \frac{n!}{r!(n-r)!}\] Here, n represents the total number of possible selections (teachers) and r represents the number of partitions (schools). However, the combination formula will not be used directly. We will first calculate the number of partitions using stars and bars theorem that allows n objects to be partitioned into r non-negative integral parts (schools). Stars and bars theorem states that the number of ways to partition n objects into r non-negative integral parts is given by: \[C(n + r - 1, r - 1)\] In this case, n = 8 (teachers) and r = 4 (schools) Applying stars and bars theorem: Divisions = \(C(8 + 4 - 1, 4 - 1) = C(11, 3)\) Now, calculate \(C(11, 3)\) to find the number of possible divisions.

Calculate the combination for unrestricted divisions: Divisions = \(C(11, 3) = \frac{11!}{3!(11-3)!}\) Divisions = \(\frac{11!}{3!8!} = \frac{11 \times 10 \times 9}{3 \times 2 \times 1} = 165\) Therefore, there are 165 possible divisions without any restrictions.

Now, we need to find the number of divisions possible when each school must receive exactly 2 teachers. Since each school receives 2 teachers, there are no possible restrictions and partitioning. The problem becomes finding the combinations of selecting 2 teachers out of 8 for each school. This will be done for all 4 schools, and in the end, we will multiply all the combinations to find the total number of divisions possible. For 8 teachers, the number of ways to choose 2 teachers for the first school is: \(C(8, 2) = \frac{8!}{2!(8-2)!} = \frac{8 \times 7}{2 \times 1} = 28\) After assigning 2 teachers to the first school, we have 6 teachers left for the other schools. We will now select 2 teachers for the second school: \(C(6, 2) = \frac{6!}{2!(6-2)!} = \frac{6 \times 5}{2 \times 1} = 15\) Do the same process for the remaining schools: For the third school: \(C(4, 2) = \frac{4!}{2!(4-2)!} = \frac{4 \times 3}{2 \times 1} = 6\) At this point, only two teachers are left. Therefore, the fourth school has only 1 possible arrangement of accepting the remaining two teachers. The total number of divisions can be found by multiplying the combinations for each school: Total Divisions = 28 (first school) × 15 (second school) × 6 (third school) × 1 (fourth school) = 2520

There are 165 possible divisions without any restrictions, while there are 2520 divisions possible when each school must receive exactly 2 teachers.