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Chapter 1: Problem 28

If 8 new teachers are to be divided among 4 schools, how many divisions are possible? What if each school must receive 2 teachers?

Short Answer

Expert verified
In summary, there are 165 possible divisions of the 8 new teachers among the 4 schools without any restrictions. However, if each school must receive exactly 2 teachers, there are 2520 possible divisions.

Step by step solution

01

Understand the problem with no restrictions

In the first part, we are asked to find the number of ways to divide 8 teachers among 4 schools without any restrictions. In this situation, the problem is equivalent to finding the number of non-negative integer solutions of the equation \(x_1 + x_2 + x_3 + x_4 = 8\), where \(x_i\) represents the number of teachers assigned to school \(i\).
02

Use Stars and Bars method

To solve this problem, we can use the Stars and Bars method. It is a combinatorial technique used to solve problems related to distributing objects into containers. According to the method, the number of ways to distribute the objects is given by the combination of objects and separator: \(\binom{n+k-1}{k-1}\), where \(n\) represents the number of objects and \(k\) represents the number of containers. In this case, we have 8 teachers (objects) and 4 schools (containers), so we can use the formula to find the total number of divisions:
03

Calculate the number of divisions without restrictions

Now, we can apply the formula to calculate the number of divisions possible: Number of divisions without restrictions = \(\binom{8+4-1}{4-1}\) = \(\binom{11}{3}\) = \(\frac{11!}{3!(11-3)!}\) = 165 So, there are 165 divisions possible without restrictions.
04

Calculate divisions with the constraint

The second part of the exercise asks us to find the number of divisions possible if each school must receive 2 teachers. In this case, the problem is straightforward as there are no other variables to consider. The number of ways to divide the 8 teachers into 4 equal groups of 2 is given by the multinomial coefficient formula: \(\frac{n!}{n_1!n_2!...n_k!}\) where \(n\) is the total number of teachers, and \(n_i\) is the number of teachers in each group. Since each school receives 2 teachers, we have the following: Number of divisions with constraint = \(\frac{8!}{2!2!2!2!}\)
05

Calculate the number of divisions with constraints

Let's calculate the value of the coefficient: Number of divisions with constraint = \(\frac{8!}{2!2!2!2!}\) = \(\frac{40320}{2*2*2*2}\) = 2520 So, there are 2520 divisions possible if each school must receive 2 teachers.

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