We are given a set \( A = \{1, 2, 3, \dots, n\} \), and we are selecting a random mapping (function) from this set into itself. We need to find the probability that this selected function is one-to-one (injective).
Let's discuss the steps and calculations required to solve this problem:
- First, we calculate the total number of possible mappings from set \( A \) into itself. Since there are \( n \) elements in set \( A \), and each element can be mapped to any of the \( n \) elements in the set, there are \( n^n \) possible mappings. This is because there are \( n \) choices for the image of each element under the mapping.
- Next, we calculate the number of one-to-one mappings (injective functions) from \( A \) to \( A \). A function is one-to-one if each element of the domain maps to a unique element in the codomain without repetition. The number of one-to-one mappings is simply the number of bijections, which is equivalent to the number of permutations of the set \( A \). This is given by \( n! \) (factorial of \( n \)).
- The probability that a randomly selected mapping is one-to-one is then the ratio of the number of one-to-one mappings to the total number of mappings. Thus, the probability \( P \) is given by:
\[ P = \frac{\text{Number of one-to-one mappings}}{\text{Total number of mappings}} = \frac{n!}{n^n} \]
Therefore, the correct answer is \(\frac{n!}{n^n}\).
The given options included:
- \(\frac{n!}{n^{n-1}}\)
- \(\frac{n!}{n^n}\) (Correct Answer)
- \(\frac{n!}{2n^n}\)
- None of these
The choice that matches our calculation is \(\frac{n!}{n^n}\), as explained above.