<p>If <span style='font-style:italic;'>P</span>(<span style='font-style:italic;'>S</span>) denotes the set of all subsets of a given set <span style='font-style:italic;'>S</span>, then the number of one-to-one functions from the set <span style='font-style:italic;'>S</span> = {1, 2, 3} to the set <span style='font-style:italic;'>P</span>(<span style='font-style:italic;'>S</span>) is</p>
Step-by-Step Solution
Key Concept: We need to count one-to-one (injective) functions from a 3-element set to the power set of that set. The power set P(S) has 2^3 = 8 elements, and we need to select 3 distinct elements from these 8 elements where order matters.
<p><strong>Step 1: Identify the domain and codomain.</strong></p><p>Domain: S = {1, 2, 3}, which has |S| = 3 elements.</p><p>Codomain: P(S) = power set of S. Since |S| = 3, we have |P(S)| = 2³ = 8 elements.</p><p>P(S) = {∅, {1}, {2}, {3}, {1,2}, {1,3}, {2,3}, {1,2,3}}</p><p><strong>Step 2: Understand one-to-one (injective) functions.</strong></p><p>A function f: S → P(S) is one-to-one if f(1), f(2), and f(3) are all distinct elements of P(S). That is, no two different elements of S map to the same element of P(S).</p><p><strong>Step 3: Count the number of one-to-one functions.</strong></p><p>For f to be injective:</p><p>• f(1) can be any of the 8 elements in P(S): 8 choices</p><p>• f(2) can be any element in P(S) except f(1): 7 choices</p><p>• f(3) can be any element in P(S) except f(1) and f(2): 6 choices</p><p><strong>Step 4: Apply the multiplication principle.</strong></p><p>Total number of one-to-one functions = 8 × 7 × 6</p><p>= 56 × 6</p><p>= 336</p><p><strong>∴ Answer: 336</strong></p>
Correct Answer: 336