Question

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

• A. 20
• B. 22
• C. 19
• D. 21

Answer 1

The Correct Option is A

Let's define:

• A = Set of students choosing mathematics
• B = Set of students choosing physics
• C = Set of students choosing chemistry

We know that:

• |A ∩ B ∩ C| = 18 (students choosing all three subjects)
• |A| = 23 (students choosing mathematics)
• |B| = 25 (students choosing physics)

We need to find the smallest possible number of students choosing chemistry (|C|). Using the principle of inclusion-exclusion for the three sets, we have:

|A ∪ B ∪ C| = |A| + |B| + |C| - |A ∩ B| - |B ∩ C| - |C ∩ A| + |A ∩ B ∩ C|

Since students must choose at least two subjects, every student is counted at least once in the pairs A ∩ B, B ∩ C, or C ∩ A, implying:

|A| + |B| + |C| - |A ∩ B| - |B ∩ C| - |C ∩ A| = 0

From here, we deduce:

|A ∪ B ∪ C| = |A| + |B| + |C| - 18≥ |A| + |B| + |C| - |A ∩ B| - |B ∩ C| - |C ∩ A| - 18

Given |A| = 23, |B| = 25:

23 + 25 + |C| - |A ∩ B| - ((23 − |A ∩ C|) + (25 − |B ∩ C|)) + 18 = 0

Rearranging gives |C|:

|C| ≥ (|A ∩ C| + |B ∩ C| + 23 + 25 - 18)

To minimize |C|, assume maximum intersection overlap. This occurs when:

Intersection	Count
|A ∩ B|	23
|B ∩ C|	25
|A ∩ C|	?

To satisfy the condition that each must choose at least two subjects:

|C| = (18 + (25 - 18) + (23 - 18)) = 20

Thus, the smallest possible number of students choosing chemistry as one of their subjects is 20.