TWELVE TEACHERS ARE TO BE ASSIGNED TO THREE SCHOOLS. In how many ways can this be done if: (a) a specific school(School 1) receives at least 2 teachers
We can approach this problem using combinations.
First, we choose 2 teachers to be assigned to School 1. This can be done in C(12,2) ways.
Next, we need to assign the remaining 10 teachers to the three schools. One possible way to do this is to use a stars and bars approach. Imagine lining up the 10 teachers in a row, and placing 2 separators (represented by bars) to divide them into 3 groups of teachers for each school. For example:
|T T T T T|T T|T T T
This represents 5 teachers for School 1, 2 teachers for School 2, and 3 teachers for School 3.
There are 9 spaces between the teachers (including the ends) where we can place the 2 bars. This can be done in C(9,2) ways.
Therefore, the total number of ways to assign 12 teachers to 3 schools such that School 1 receives at least 2 teachers is:
C(12,2) x C(9,2) = 66 x 36 = 2376 ways.
To find the number of ways to assign twelve teachers to three schools, we can consider different cases based on the minimum number of teachers assigned to School 1.
Case 1: School 1 receives 2 teachers:
In this case, we can select 2 teachers out of 12 to be assigned to School 1 in 12C2 ways. The remaining 10 teachers can be assigned to the remaining 2 schools in 2^10 ways (each teacher can go to either School 2 or School 3). Therefore, the number of ways for this case is 12C2 * 2^10.
Case 2: School 1 receives 3 teachers:
We can select 3 teachers out of 12 for School 1 in 12C3 ways. The remaining 9 teachers can be assigned to the remaining 2 schools in 2^9 ways. So the number of ways for this case is 12C3 * 2^9.
Case 3: School 1 receives 4 teachers:
We can select 4 teachers out of 12 for School 1 in 12C4 ways. The remaining 8 teachers can be assigned to the remaining 2 schools in 2^8 ways. So the number of ways for this case is 12C4 * 2^8.
Case 4: School 1 receives 5 teachers:
We can select 5 teachers out of 12 for School 1 in 12C5 ways. The remaining 7 teachers can be assigned to the remaining 2 schools in 2^7 ways. So the number of ways for this case is 12C5 * 2^7.
Now, we need to sum up the number of ways for all these cases to get the final answer:
Total number of ways = 12C2 * 2^10 + 12C3 * 2^9 + 12C4 * 2^8 + 12C5 * 2^7.
Depending on the specific values of the binomial coefficients (12C2, 12C3, etc.), you can calculate each term and then add them up to find the total number of ways.