Question

Let \(S = \{1, 2, 3, 4, 5, 6\}\). Then the probability that a randomly chosen onto function \(g\) from \(S\) to \(S\) satisfies \(g(3) = 2g(1)\) is :

• A. \(\frac{1}{30}\)
• B. \(\frac{1}{15}\)
• C. \(\frac{1}{10}\)
• D. \(\frac{1}{5}\)

Hint:

For a function \(f: A \to A\) where \(|A| = n\), the term "onto" implies the function is a bijection. Always identify such properties first to simplify the counting process in probability.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
Since function \(g\) is from \(S\) to \(S\) and is "onto", and the sets have the same number of elements (6), the function must also be one-to-one (a bijection). This means \(g\) is a permutation of the elements of \(S\).
Step 2: Key Formula or Approach:
1. Total number of onto functions from \(S\) to \(S\) is \(n! = 6!\).
2. Favorable outcomes are permutations where the specific condition \(g(3) = 2g(1)\) is satisfied.
Step 3: Detailed Explanation:
The set is \(S = \{1, 2, 3, 4, 5, 6\}\).
The condition is \(g(3) = 2g(1)\). Since the range of \(g\) is also \(S\), the possible values for \(g(1)\) and \(g(3)\) are:
- Case 1: \(g(1) = 1, g(3) = 2\)
- Case 2: \(g(1) = 2, g(3) = 4\)
- Case 3: \(g(1) = 3, g(3) = 6\)
For each case, we have fixed the mapping for 2 elements of the domain (\(1\) and \(3\)). Since the function must be a bijection, the remaining \(6 - 2 = 4\) elements in the domain (\(\{2, 4, 5, 6\}\)) can be mapped to the remaining 4 elements in the codomain in \(4!\) ways.
Total favorable cases = \(3 \times 4! = 3 \times 24 = 72\).
Total onto functions = \(6! = 720\).
Probability = \(\frac{\text{Favorable cases}}{\text{Total cases}} = \frac{72}{720} = \frac{1}{10}\).
Step 4: Final Answer:
The probability is \(\frac{1}{10}\).