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Difficulty: Very hardSets and Set Operations

In a survey of 120 final year secondary school students in Lagos, 47 offer Physics (PP), 57 offer Chemistry (CC), and 48 offer Biology (BB). Records show that 17 students offer both Physics and Chemistry, 20 offer both Chemistry and Biology, and 15 offer both Physics and Biology. If the number of students who offer none of these three subjects is three times the number of students who offer all three subjects, find the total number of students who offer exactly one of the three subjects.

Answer: 63 students

Answer

63 students offer exactly one of the three subjects.
By applying the principle of inclusion-exclusion for three sets, the total union is ∣P∪C∪B∣=47+57+48−(17+20+15)+x=100+x|P \cup C \cup B| = 47 + 57 + 48 - (17 + 20 + 15) + x = 100 + x, where xx is the number of students taking all three subjects. Setting the universal set total to 120 gives (100+x)+3x=120(100 + x) + 3x = 120, solving to x=5x = 5. Subtracting the relevant intersections yields 20 students for Physics only, 25 for Chemistry only, and 18 for Biology only. Summing these gives 63.

Step-by-Step Solution

1
Define unknown variables and set up the inclusion-exclusion formula for the union of three sets.
∣P∪C∪B∣=100+x|P \cup C \cup B| = 100 + x, where x=∣P∩C∩B∣x = |P \cap C \cap B|.
The Principle of Inclusion-Exclusion states that ∣P∪C∪B∣=∣P∣+∣C∣+∣B∣−(∣P∩C∣+∣C∩B∣+∣P∩B∣)+∣P∩C∩B∣|P \cup C \cup B| = |P| + |C| + |B| - (|P \cap C| + |C \cap B| + |P \cap B|) + |P \cap C \cap B|.
2
Formulate and solve an equation for the total number of students in the universal set.
120=(100+x)+3x  ⟹  4x=20  ⟹  x=5120 = (100 + x) + 3x \implies 4x = 20 \implies x = 5.
The total number of students equals those taking at least one subject plus those taking none (3x3x).
3
Determine the number of students taking exactly two subjects.
Physics and Chemistry only = 17−5=1217 - 5 = 12; Chemistry and Biology only = 20−5=1520 - 5 = 15; Physics and Biology only = 15−5=1015 - 5 = 10.
Subtract the triple intersection count (x=5x = 5) from each pairwise intersection count.
4
Calculate the number of students offering only Physics, only Chemistry, and only Biology.
Physics only = 47−(12+10+5)=2047 - (12 + 10 + 5) = 20; Chemistry only = 57−(12+15+5)=2557 - (12 + 15 + 5) = 25; Biology only = 48−(10+15+5)=1848 - (10 + 15 + 5) = 18.
Subtract the sum of the two-subject-only regions and the three-subject region from each set's total cardinality.
5
Sum the counts of students offering exactly one subject.
20+25+18=6320 + 25 + 18 = 63.
The total offering exactly one subject is the sum of the three disjoint single-subject regions.

Key Concept

3-Set Principle of Inclusion-Exclusion and Venn Diagram Region Decomposition
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Topics

Sets and Set Operations

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