In a survey conducted among 300 State PSC aspirants, data was collected regarding their en… — State PSC Exam

In a survey conducted among $300$ State PSC aspirants, data was collected regarding their enrollment in preparation modules for three optional subjects: Public Administration ( $P$ ), Sociology ( $S$ ), and Geography ( $G$ ). It was found that $140$ aspirants enrolled in Public Administration, $130$ in Sociology, and $120$ in Geography. Further, $50$ aspirants enrolled in both Public Administration and Sociology, $45$ in both Sociology and Geography, and $40$ in both Public Administration and Geography. If $20$ aspirants did not enroll in any of these three subjects, how many aspirants enrolled in EXACTLY ONE optional subject?

195Answer
B
120
C
145
D
165

Answer

The number of aspirants enrolled in exactly one optional subject is 195.

By applying the inclusion-exclusion principle for three overlapping sets, the number of aspirants enrolled in all three subjects is determined to be

25

. Decomposing each set into disjoint regions gives

75

candidates in Public Administration only,

60

in Sociology only, and

60

in Geography only. Summing these exclusive single-set regions yields

195

Step-by-Step Solution

Determine the total number of aspirants enrolled in at least one optional subject

Total enrolled

|P \cup S \cup G| = 300 - 20 = 280

Candidates not enrolled in any of the three subjects must be excluded from the universe of

300

Apply the Principle of Inclusion-Exclusion for three sets to find the three-set intersection region

|P \cap S \cap G|

|P \cup S \cup G| = |P| + |S| + |G| - (|P \cap S| + |S \cap G| + |P \cap G|) + |P \cap S \cap G| \implies 280 = 140 + 130 + 120 - (50 + 45 + 40) + |P \cap S \cap G| \implies 280 = 390 - 135 + |P \cap S \cap G| \implies |P \cap S \cap G| = 25

To find individual exclusive regions, the central intersection of all three sets must first be determined.

Calculate the count of candidates enrolled in exactly two subjects

Only

(P \cap S) = 50 - 25 = 25

; Only

(S \cap G) = 45 - 25 = 20

; Only

(P \cap G) = 40 - 25 = 15

Subtracting the triple intersection from pairwise intersections gives the exact counts of elements belonging strictly to two sets.

Calculate the count of candidates enrolled in exactly one subject for each subject and sum them up

Only

P = 140 - (25 + 25 + 15) = 75

; Only

S = 130 - (25 + 25 + 20) = 60

; Only

G = 120 - (15 + 25 + 20) = 60

. Total exactly one =

75 + 60 + 60 = 195

Subtracting all overlapping regions from each subject set total yields the number of candidates taking only that single subject.

Key Concept

Principle of Inclusion-Exclusion for Three Sets

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