In a department of 70 university lecturers, 40 publish research in Journal A, 30 publish i… — JAMB UTME

In a department of $70$ university lecturers, $40$ publish research in Journal $A$ , $30$ publish in Journal $B$ , and $25$ publish in Journal $C$ . It is known that $15$ publish in both Journals $A$ and $B$ , $12$ publish in both Journals $B$ and $C$ , $10$ publish in both Journals $A$ and $C$ , and $5$ publish in all three journals. How many lecturers do not publish in any of these three journals?

$7$ Answer
B
$12$
C
$17$
D
$63$

Answer

The number of lecturers who do not publish in any of the three journals is

7

Using the inclusion-exclusion principle for three sets,

n(A \cup B \cup C) = 40 + 30 + 25 - 15 - 12 - 10 + 5 = 63

. The number of lecturers publishing in none of the journals is the complement of this union relative to the universal set of

70

, which is

70 - 63 = 7

Step-by-Step Solution

Apply the Principle of Inclusion-Exclusion for three sets to find the total number of lecturers who publish in at least one journal,

n(A \cup B \cup C)

n(A \cup B \cup C) = n(A) + n(B) + n(C) - [n(A \cap B) + n(B \cap C) + n(A \cap C)] + n(A \cap B \cap C)

Elements counted multiple times in pairwise intersections must be subtracted, and the central triple intersection must be added back.

Substitute the given numerical values into the formula.

n(A \cup B \cup C) = 40 + 30 + 25 - (15 + 12 + 10) + 5 = 95 - 37 + 5 = 63

To evaluate the total cardinality of the union.

Subtract

n(A \cup B \cup C)

from the universal set size

n(U)

n((A \cup B \cup C)') = n(U) - n(A \cup B \cup C) = 70 - 63 = 7

The number of lecturers publishing in none of the journals corresponds to the complement of the union of all three sets.

Key Concept

Principle of Inclusion-Exclusion for Three Sets and Set Complement

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