Question:
A mapping is selected at random from all mappings \(f:A\rightarrow A\) where set \(A=\{1,2,3,\dots,n\}\). If the probability that the mapping is injective is \(\frac{3}{32}\), then the value of \(n\) is:
A mapping is selected at random from all mappings \(f:A\rightarrow A\) where set \(A=\{1,2,3,\dots,n\}\). If the probability that the mapping is injective is \(\frac{3}{32}\), then the value of \(n\) is:
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For function counting problems, remember: Total functions $= (\text{Size of Codomain})^{(\text{Size of Domain})}$, while One-to-One functions $= {}^{(\text{Codomain})}P_{(\text{Domain})}$. If the domain is larger than the codomain, the number of one-to-one functions drops to 0 immediately!
Updated On: May 28, 2026
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The Correct Option is D
Solution and Explanation
Concept:
The probability of a random event is defined as the number of favorable configurations divided by the total size of the sample space. For functions mapping a finite set to itself, we can use permutations and exponent rules to count these configurations.
Step 1: Count the total number of possible mappings.
The function maps a domain set $A$ containing $n$ elements into a codomain set $A$ which also contains $n$ elements. Each independent element in the domain has $n$ possible target options to map to. Therefore, the total number of unique functions that can be created is: $$\text{Total Mappings } (N) = n \times n \times \dots \times n = n^n$$
Step 2: Count the total number of injective functions.
An injective (one-to-one) function requires that no two elements in the domain map to the same target element in the codomain.
• The first element has $n$ choices available.
• The second element can choose from $(n-1)$ remaining options.
• The $n$-th element is left with exactly $1$ choice. This matches the standard definition of a factorial permutation: $$\text{Injective Mappings } (n_f) = n \times (n-1) \times \dots \times 1 = n!$$
Step 3: Set up the probability equation and solve for $n$.
The probability of selecting an injective function at random is: $$P(n) = \frac{\text{Injective Mappings}}{\text{Total Mappings}} = \frac{n!}{n^n} = \frac{3}{32}$$ Since $n$ must be a positive integer, let us test the values from our answer choices to find a match:
• Test $n = 3$: $P(3) = \frac{3!}{3^3} = \frac{6}{27} = \frac{2}{9} \neq \frac{3}{32}$
• Test $n = 4$: $P(4) = \frac{4!}{4^4} = \frac{24}{256} = \frac{3}{32}$ The value $n = 4$ satisfies the target probability ratio perfectly, confirming option (D).
The function maps a domain set $A$ containing $n$ elements into a codomain set $A$ which also contains $n$ elements. Each independent element in the domain has $n$ possible target options to map to. Therefore, the total number of unique functions that can be created is: $$\text{Total Mappings } (N) = n \times n \times \dots \times n = n^n$$
Step 2: Count the total number of injective functions.
An injective (one-to-one) function requires that no two elements in the domain map to the same target element in the codomain.
• The first element has $n$ choices available.
• The second element can choose from $(n-1)$ remaining options.
• The $n$-th element is left with exactly $1$ choice. This matches the standard definition of a factorial permutation: $$\text{Injective Mappings } (n_f) = n \times (n-1) \times \dots \times 1 = n!$$
Step 3: Set up the probability equation and solve for $n$.
The probability of selecting an injective function at random is: $$P(n) = \frac{\text{Injective Mappings}}{\text{Total Mappings}} = \frac{n!}{n^n} = \frac{3}{32}$$ Since $n$ must be a positive integer, let us test the values from our answer choices to find a match:
• Test $n = 3$: $P(3) = \frac{3!}{3^3} = \frac{6}{27} = \frac{2}{9} \neq \frac{3}{32}$
• Test $n = 4$: $P(4) = \frac{4!}{4^4} = \frac{24}{256} = \frac{3}{32}$ The value $n = 4$ satisfies the target probability ratio perfectly, confirming option (D).
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