Question:
Let $A$ and $B$ be two non-empty sets with $n(A)=4$ and $n(B)=5$. If a mapping is selected at random from the set of all mappings from $A$ to $B$, then the probability of getting a many-one mapping is
Let $A$ and $B$ be two non-empty sets with $n(A)=4$ and $n(B)=5$. If a mapping is selected at random from the set of all mappings from $A$ to $B$, then the probability of getting a many-one mapping is
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Many-one mappings = Total mappings $-$ One-one mappings.
Updated On: Jun 3, 2026
- $\dfrac{29}{125}$
- $\dfrac{24}{125}$
- $\dfrac{96}{125}$
- $\dfrac{101}{125}$
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The Correct Option is D
Solution and Explanation
Step 1: Concept
Probability = $\dfrac{\text{Number of many-one mappings}}{\text{Total mappings}}$.
Step 2: Meaning
Total mappings from $A$ to $B$: \[ 5^4=625. \]
Step 3: Analysis
One-one mappings from a $4$-element set to a $5$-element set are \[ {}^{5}P_{4} = 5\cdot4\cdot3\cdot2 = 120. \] Hence many-one mappings are \[ 625-120=505. \] Therefore \[ P = \frac{505}{625} = \frac{101}{125}. \]
Step 4: Conclusion
Thus the probability of obtaining a many-one mapping is \[ \frac{101}{125}. \]
Final Answer: (D)
Probability = $\dfrac{\text{Number of many-one mappings}}{\text{Total mappings}}$.
Step 2: Meaning
Total mappings from $A$ to $B$: \[ 5^4=625. \]
Step 3: Analysis
One-one mappings from a $4$-element set to a $5$-element set are \[ {}^{5}P_{4} = 5\cdot4\cdot3\cdot2 = 120. \] Hence many-one mappings are \[ 625-120=505. \] Therefore \[ P = \frac{505}{625} = \frac{101}{125}. \]
Step 4: Conclusion
Thus the probability of obtaining a many-one mapping is \[ \frac{101}{125}. \]
Final Answer: (D)
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