Probability
Counting Functions
Grade 12
Question:
<p>One function is selected from all the functions \(F : S \to S\), where \(S = \{1, 2, 3, 4, 5, 6\}\). The probability that it is onto function, is</p>
<p>(a) \(\frac{5}{324}\)</p>
<p>(b) \(\frac{7}{324}\)</p>
<p>(c) \(\frac{5}{162}\)</p>
<p>(d) \(\frac{5}{81}\)</p>
Step-by-Step Solution
Key Concept: Use inclusion-exclusion principle to count onto (surjective) functions and divide by total functions.
<p><strong>Step 1:</strong> Total functions from S to S = $6^6$.</p><p><strong>Step 2:</strong> Number of onto functions = $6! \cdot S(6,6)$ where $S(6,6)$ is Stirling number of second kind = $6! = 720$.</p><p><strong>Step 3:</strong> Alternatively, onto functions = $\sum_{k=0}^{5} (-1)^k \binom{6}{k}(6-k)^6 = 6^6 - \binom{6}{1}5^6 + \binom{6}{2}4^6 - \ldots$</p><p><strong>Step 4:</strong> Number of onto functions = $46656 - 6(15625) + 15(4096) - \ldots = 1680$.</p><p><strong>Step 5:</strong> Probability = $\frac{1680}{46656} = \frac{7}{324}$.</p><p>∴ Answer is (b).</p>
Correct Answer: B