Probability
Counting Functions
Grade 12
Question:
<p>Let \(S\) be the set of all functions from the set \(\{1, 2, \ldots, 10\}\) to itself. One function is selected from \(S\), the probability that the selected function is one-one and onto is:</p>
<p>(a) \(\frac{9!}{10^9}\)</p>
<p>(b) \(\frac{1}{10}\)</p>
<p>(c) \(\frac{100}{10!}\)</p>
<p>(d) \(\frac{9!}{10^{10}}\)</p>
Step-by-Step Solution
Key Concept: A function from a finite set to itself is one-one and onto if and only if it is a bijection/permutation. Count bijections versus total functions.
<p>Total functions from a set of 10 elements to itself: \(10^{10}\)</p><p>One-one and onto functions (bijections/permutations) from a 10-element set to itself: \(10!\)</p><p>Probability = \(\frac{10!}{10^{10}} = \frac{10 \times 9!}{10^{10}} = \frac{9!}{10^9}\)</p>
Correct Answer: A