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
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
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For one-one mapping, each element in codomain gets at most one preimage.
Updated On: Apr 7, 2026
- \(\frac{n!}{n^{n-1}}\)
- \(\frac{n!}{n^n}\)
- \(\frac{n!}{2n^n}\)
- None of these
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Concept:
Total mappings = \(n^n\), one-one mappings = \(n!\).
Step 2: Detailed Explanation:
Total number of functions from \(A\) to \(A\) = \(n^n\)
Number of injective (one-one) functions = \(n!\)
Probability = \(\frac{n!}{n^n}\)
Step 3: Final Answer:
\(\frac{n!}{n^n}\).
Total mappings = \(n^n\), one-one mappings = \(n!\).
Step 2: Detailed Explanation:
Total number of functions from \(A\) to \(A\) = \(n^n\)
Number of injective (one-one) functions = \(n!\)
Probability = \(\frac{n!}{n^n}\)
Step 3: Final Answer:
\(\frac{n!}{n^n}\).
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