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!
Step 1: Understanding the Concept:
A mapping from a set with \( n \) elements to itself is simply a function where each element of the domain has \( n \) possible choices in the codomain. An "injective" (one-to-one) mapping means every element in the domain maps to a unique element in the codomain. Step 2: Key Formula or Approach:
1. Total number of mappings: \( n^n \).
2. Total number of injective mappings: \( n! \) (Permutation of \( n \) elements).
3. Probability \( P = \frac{n!}{n^n} \). Step 3: Detailed Explanation:
We are given that \( P = \frac{3}{32} \). So:
\[ \frac{n!}{n^n} = \frac{3}{32} \]
Let's test the integer options provided:
- If \( n = 3 \):
\[ \frac{3!}{3^3} = \frac{6}{27} = \frac{2}{9} \approx 0.222 \]
\[ \frac{3}{32} \approx 0.093 \]
This is not a match.
- If \( n = 4 \):
\[ \frac{4!}{4^4} = \frac{24}{256} \]
Divide numerator and denominator by 8:
\[ 24 \div 8 = 3 \]
\[ 256 \div 8 = 32 \]
The result is exactly \( \frac{3}{32} \). Step 4: Final Answer:
The value of \( n \) that satisfies the probability condition is 4.