Step-by-Step Deconstruction
Topic: Relations and Functions Mapping
(i) Total Number of Relations
Number of elements in S, n(S) = 4. Number of elements in J, n(J) = 3.
Total relations = 2n(S) × n(J) = 24 × 3 = 212 = 4096.
(ii) Bijective Check for f
f = {(S₁, J₁), (S₂, J₂), (S₃, J₂), (S₄, J₃)}
Not Bijective: For a function to be bijective, it must be both One-One and Onto.
- Not One-One: S₂ and S₃ both map to J₂ (f(S₂) = f(S₃) = J₂).
(iii) (a) Number of One-One Functions
For a one-one function from S to J, we need n(S) ≤ n(J).
Here, n(S) = 4 and n(J) = 3.
Since the number of speakers is greater than the number of judges, at least two speakers must share a judge (Pigeonhole Principle). Therefore, total one-one functions = 0.
(iii) (b) Equivalence Building in R₁
R₁ = {(S₁, S₂), (S₂, S₄)} defined in set S = {S₁, S₂, S₃, S₄}.
Reflexive Requirements: Every element must relate to itself. Add (S₁, S₁), (S₂, S₂), (S₃, S₃), (S₄, S₄).
Symmetric Constraint: The question asks to remain Not Symmetric. Since (S₁, S₂) is present but (S₂, S₁) is not, it is already not symmetric. No further pairs required to break symmetry.
⚠️ Silly Mistake Audit: The Domain Trap
In part (iii)(a), many students try to use the nPr formula (3P4), which results in an error. Always check the sizes first: if the Domain is larger than the Co-domain, a One-One function is mathematically impossible!