Home Class 11 Applied Maths Chapter 4: Sets Exercise 4.5: Cardinality and Practical Applications
Chapter 4 • Unit II

Exercise 4.5: Cardinality and Practical Applications

Comprehensive guide and practice questions for CBSE Class 11 Applied Mathematics curriculum (2026-27).

4.5 Cardinality and Practical Applications of Sets

Cardinality — the count of elements in a finite set — is the quantitative bridge connecting pure set algebra to real-world business analytics, consumer surveys, market research, and risk modeling. The Principle of Inclusion-Exclusion (PIE) enables analysts to prevent double-counting across overlapping customer demographics or product lines.

Fundamental Cardinality Theorems & Formulae

1. Two-Set System (\(A\) and \(B\) within Universal Set \(U\))

  • Union of Two Sets: \[ n(A \cup B) = n(A) + n(B) - n(A \cap B) \]
  • Disjoint Sets: If \(A \cap B = \emptyset\), then \(n(A \cup B) = n(A) + n(B)\).
  • Only \(A\) (Difference \(A - B\) or \(A \cap B'\)): \[ n(A - B) = n(A) - n(A \cap B) = n(A \cup B) - n(B) \]
  • Symmetric Difference \(A \Delta B\) (Exactly one of \(A\) or \(B\)): \[ n(A \Delta B) = n(A - B) + n(B - A) = n(A) + n(B) - 2n(A \cap B) = n(A \cup B) - n(A \cap B) \]
  • Neither \(A\) nor \(B\) (De Morgan's Complement): \[ n(A' \cap B') = n((A \cup B)') = n(U) - n(A \cup B) \]

2. Three-Set System (\(A, B, C\) within Universal Set \(U\))

  • Union of Three Sets: \[ n(A \cup B \cup C) = n(A) + n(B) + n(C) - n(A \cap B) - n(B \cap C) - n(C \cap A) + n(A \cap B \cap C) \]
  • Elements in Exactly Two of the Three Sets: \[ n(\text{Exactly 2}) = [n(A \cap B) + n(B \cap C) + n(C \cap A)] - 3n(A \cap B \cap C) \]
  • Elements in at Least Two Sets: \[ n(\text{At least 2}) = [n(A \cap B) + n(B \cap C) + n(C \cap A)] - 2n(A \cap B \cap C) \]
  • Elements in Exactly One Set: \[ n(\text{Exactly 1}) = n(A) + n(B) + n(C) - 2[n(A \cap B) + n(B \cap C) + n(C \cap A)] + 3n(A \cap B \cap C) \]

Systematic 4-Step Region Filling Strategy for Survey Problems

When solving complex market research problems, setting up algebraic Venn diagram regions prevents algebraic errors:

  1. Anchor at the Core: Always start by labeling or determining the innermost intersection of all sets \(n(A \cap B \cap C) = x\).
  2. Determine Pairwise Exclusives: Subtract \(x\) from pairwise intersections to find elements belonging to only two sets: \[ n(A \cap B \text{ only}) = n(A \cap B) - x \]
  3. Determine Singleton Exclusives: Subtract the three adjoining intersection regions from the individual set totals to find elements in only one set: \[ n(A \text{ only}) = n(A) - [n(A \cap B \text{ only}) + n(A \cap C \text{ only}) + x] \]
  4. Account for Non-Participants (Outside): Subtract the union \(n(A \cup B \cup C)\) from the universal set \(n(U)\) to find elements satisfying none of the conditions.

Comprehensive Applied Example: Fintech Consumer Survey

Scenario: A market research agency surveys \(1{,}000\) smartphone users regarding their usage of three digital payment apps: PhonePe (\(P\)), Google Pay (\(G\)), and Paytm (\(T\)). The survey reports:

  • \(n(P) = 450\), \(n(G) = 380\), \(n(T) = 280\)
  • \(n(P \cap G) = 160\), \(n(G \cap T) = 120\), \(n(P \cap T) = 110\)
  • \(n(P \cap G \cap T) = 50\) (use all three apps)

Questions to Analyze:

  1. How many users use at least one digital payment app?
  2. How many users use none of these three apps?
  3. How many users use exactly two apps?
  4. How many users use PhonePe only?

Step-by-Step Solution:

  1. Users using at least one app: \[ n(P \cup G \cup T) = n(P) + n(G) + n(T) - [n(P \cap G) + n(G \cap T) + n(P \cap T)] + n(P \cap G \cap T) \] \[ n(P \cup G \cup T) = 450 + 380 + 280 - (160 + 120 + 110) + 50 \] \[ n(P \cup G \cup T) = 1110 - 390 + 50 = 770 \text{ users.} \]
  2. Users using none of the three apps: \[ n((P \cup G \cup T)') = n(U) - n(P \cup G \cup T) = 1000 - 770 = 230 \text{ users.} \]
  3. Users using exactly two apps: \[ n(\text{Exactly 2}) = [n(P \cap G) + n(G \cap T) + n(P \cap T)] - 3n(P \cap G \cap T) \] \[ n(\text{Exactly 2}) = (160 + 120 + 110) - 3(50) = 390 - 150 = 240 \text{ users.} \] Verification via pairwise exclusives:
    • \(P \cap G\) only \(= 160 - 50 = 110\)
    • \(G \cap T\) only \(= 120 - 50 = 70\)
    • \(P \cap T\) only \(= 110 - 50 = 60\)
    • Total \(= 110 + 70 + 60 = 240\). Matches perfectly.
  4. Users using PhonePe only: \[ n(P \text{ only}) = n(P) - [n(P \cap G \text{ only}) + n(P \cap T \text{ only}) + n(P \cap G \cap T)] \] \[ n(P \text{ only}) = 450 - (110 + 60 + 50) = 450 - 220 = 230 \text{ users.} \]
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