Question:medium

Students in a college have to choose at least two subjects from chemistry, mathematics and physics. The number of students choosing all three subjects is 18, choosing mathematics as one of their subjects is 23 and choosing physics as one of their subjects is 25. The smallest possible number of students who could choose chemistry as one of their subjects is

Updated On: Jan 15, 2026
  • 20
  • 22
  • 19
  • 21
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The Correct Option is A

Solution and Explanation

Let the following sets represent students choosing subjects:
  • A = Students selecting mathematics
  • B = Students selecting physics
  • C = Students selecting chemistry
Given information includes:
  • |A ∩ B ∩ C| = 18 (students choosing all three subjects)
  • |A| = 23 (students selecting mathematics)
  • |B| = 25 (students selecting physics)
The objective is to determine the minimum possible value for |C|, the number of students selecting chemistry. Applying the principle of inclusion-exclusion for three sets:
|A ∪ B ∪ C| = |A| + |B| + |C| - |A ∩ B| - |B ∩ C| - |C ∩ A| + |A ∩ B ∩ C|
Since every student must choose at least two subjects, each student is accounted for in at least one of the pairwise intersections (A ∩ B, B ∩ C, or C ∩ A). This implies:
|A| + |B| + |C| - |A ∩ B| - |B ∩ C| - |C ∩ A| = 0
Therefore:
|A ∪ B ∪ C| = |A| + |B| + |C| - 18≥ |A| + |B| + |C| - |A ∩ B| - |B ∩ C| - |C ∩ A| - 18
With |A| = 23 and |B| = 25:
23 + 25 + |C| - |A ∩ B| - ((23 − |A ∩ C|) + (25 − |B ∩ C|)) + 18 = 0
Rearranging to solve for |C|:
|C| ≥ (|A ∩ C| + |B ∩ C| + 23 + 25 - 18)
To minimize |C|, we assume maximum overlap within the intersections. This occurs when:
IntersectionCount
|A ∩ B|23
|B ∩ C|25
|A ∩ C|?
To fulfill the condition that each student selects at least two subjects:
|C| = (18 + (25 - 18) + (23 - 18)) = 20
Consequently, the minimum number of students selecting chemistry is 20.
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