In a group of 100 candidates preparing for an entrance examination, 48 registered for Math… — JAMB UTME

In a group of 100 candidates preparing for an entrance examination, 48 registered for Mathematics, 45 for Physics, and 40 for Chemistry. If 18 candidates registered for both Mathematics and Physics, 15 for both Physics and Chemistry, 20 for both Mathematics and Chemistry, and 8 registered for none of these three subjects, how many candidates registered for all three subjects?

Answer: 12 / 12 candidates / 12 students

Answer

12 candidates registered for all three subjects.

By the Principle of Inclusion-Exclusion, the total number of candidates taking at least one subject is

100 - 8 = 92

. Summing the individual totals gives

48 + 45 + 40 = 133

. Subtracting the pairwise intersections gives

133 - (18 + 15 + 20) = 80

. Adding the intersection of all three sets must equal 92, yielding

92 - 80 = 12

Step-by-Step Solution

Determine the total number of candidates who registered for at least one of the three subjects.

n(M \cup P \cup C) = 100 - 8 = 92

Subtracting the number of candidates who registered for none of the subjects from the universal set gives the cardinality of the union.

Apply the Principle of Inclusion-Exclusion for three sets.

n(M \cup P \cup C) = n(M) + n(P) + n(C) - n(M \cap P) - n(P \cap C) - n(M \cap C) + n(M \cap P \cap C)

This formula relates the individual set sizes, pair intersections, and triple intersection to the union.

Substitute the known values into the inclusion-exclusion equation.

92 = 48 + 45 + 40 - 18 - 15 - 20 + n(M \cap P \cap C)

Insert the given cardinalities into the formula.

Simplify and solve for

n(M \cap P \cap C)

92 = 80 + n(M \cap P \cap C) \implies n(M \cap P \cap C) = 92 - 80 = 12

Isolate the unknown triple intersection term.

Key Concept

Principle of Inclusion-Exclusion for Three Sets

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