Question

One mapping (function) is selected at random from all the mappings of the set \( A = \{1,2,3,\dots,n\} \) into itself. The probability that the mapping selected is one-one, is

• A. \( \frac{n!}{n^{n-1}} \)
• B. \( \frac{n!}{n^n} \)
• C. \( \frac{n!}{2n^n} \)
• D. None of the above

Hint:

Total functions from a set of size \( n \) to itself is \( n^n \), and bijections are \( n! \).

Answer 1

The Correct Option is B

To solve the problem of finding the probability that a randomly selected mapping from the set \( A = \{1, 2, 3, \dots, n\} \) into itself is a one-one function, we proceed as follows:

1. Understand the Definition:
   • A one-one function (injective function) from a set \(A\) into itself means that each element of the set \(A\) maps to a unique element of the set, without any repetitions.
2. Total Number of Mappings:
   • For a set \(A\) with \(n\) elements mapping into itself, each element can be mapped to any of the \(n\) elements. Thus, the total number of possible mappings is \(n^n\).
3. Number of One-One Mappings:
   • To form a one-one mapping from \(A\) into itself, each element must map to a different element. Choosing a different element for each position without repetition results in a permutation.
   • Therefore, the number of one-one mappings is \(n!\), which is the number of permutations of \(n\) elements.
4. Calculate the Probability:
   • The probability that a randomly selected mapping is one-one can be calculated by dividing the number of one-one mappings by the total number of mappings:
   • \(\text{Probability} = \frac{n!}{n^n}\)
5. Conclusion:
   • Thus, the probability that the mapping selected is one-one is \(\frac{n!}{n^n}\), which corresponds to the option: \(\frac{n!}{n^n}\).