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Permutations refer to the different ways in which a set of items can be arranged in order. In our exercise, permutation plays a crucial role as it determines the number of ways the professor can distribute the four textbooks among the students. Since each textbook must go to one student and there are four textbooks, we calculate the total number of permutations for these four items using the factorial of 4, denoted as \(4!\). This is computed as follows:

\[4! = 4 \times 3 \times 2 \times 1 = 24\]

Thus, there are 24 possible ways to distribute the books randomly. Understanding permutations allows us to organize and count these arrangements efficiently.

Combinatorics is the field of mathematics that deals with counting, both as a means and an end in obtaining results. In the problem at hand, we use combinatorics to solve for various probability scenarios.

**List Possible Outcomes**
To solve part of the exercise, we first need to acknowledge the total number of possible ways to distribute the books, which involves listing or considering all permutations of the textbooks among students.

**Choosing Outcomes**
For specific questions such as when exactly two students receive their correct book, combinatorics helps us calculate this using combinations. We select two students from a group of four who will definitely receive their own books. This is indicated by the combination \({4 \choose 2}\), representing how many subsets of 2 we can form, which is calculated as 6. Then, we account for the arrangement of the other unmapped students' books, leading to a total of \(6 \times 2 = 12\) favorable outcomes.

Combinatorics provides the tools needed for organizing and counting these outcomes to identify probabilities.

In probability theory, equally likely outcomes refer to scenarios where each outcome in the sample space has the same chance of occurring. This assumption is crucial for calculating probabilities.

**Equally Likely in Textbook Distribution**
In our exercise, every possible permutation of the textbooks is considered equally likely. This implies that since there are 24 permutations, the probability of any specific permutation occurring, such as a student receiving their own book, is \(\frac{1}{24}\).

By understanding that each outcome has an equal chance, we simplify the calculation of probabilities. For example, the probability that exactly two students receive their own books is obtained by dividing the number of favorable outcomes (12) by the total number of outcomes (24), resulting in \(\frac{1}{2}\).

This concept is foundational in probability and helps ensure fairness and accuracy in our calculations when each arrangement or case is assumed to be equally probable.