A group of 40 students appeared in an examination of 3 subjects – Mathematics, Physics and Chemistry. It was found that all students passed in atleast one of the subjects, 20 students passed in Mathematics, 25 students passed in Physics, 16 students passed in Chemistry, atmost 11 students passed in both Mathematics and Physics, atmost 15 students passed in both Physics and Chemistry, atmost 15 students passed in both Mathematics and Chemistry. The maximum number of students passed in all the three subjects is _____

The correct answer is 10.

We are given the following information about 40 students and 3 subjects — Mathematics (M), Physics (P), and Chemistry (C):

|M| = 20,   |P| = 25,   |C| = 16
Total students = 40, and every student passed in at least one subject.
|M ∩ P| ≤ 11,   |P ∩ C| ≤ 15,   |M ∩ C| ≤ 15

We want to find the maximum number of students who passed in all three subjects, i.e., maximize |M ∩ P ∩ C|.

Step 1: Apply the Inclusion-Exclusion Principle.

By the inclusion-exclusion principle for three sets:

|M\cup P\cup C|=\left|M\right|+\left|P\right|+\left|C\right|-|M\cap P|-|P\cap C|-|M\cap C|+|M\cap P\cap C|

Since all 40 students passed in at least one subject:

|M\cup P\cup C|=40

Substituting:

40=20+25+16-|M\cap P|-|P\cap C|-|M\cap C|+|M\cap P\cap C|

40=61-\left(\right|M\cap P|+|P\cap C|+|M\cap C\left|\right)+|M\cap P\cap C|

Step 2: Rearrange to isolate |M ∩ P ∩ C|.

|M\cap P\cap C|=40-61+\left(\right|M\cap P|+|P\cap C|+|M\cap C\left|\right)

|M\cap P\cap C|=-21+\left(\right|M\cap P|+|P\cap C|+|M\cap C\left|\right)

Step 3: To maximize |M ∩ P ∩ C|, maximize the pairwise intersections.

Since the problem says at most for each pairwise intersection, we use their maximum allowed values:

|M ∩ P| ≤ 11  ⟹ use 11
|P ∩ C| ≤ 15  ⟹ use 15
|M ∩ C| ≤ 15  ⟹ use 15

|M\cap P\cap C|=-21+(11+15+15)

|M\cap P\cap C|=-21+41=20

Step 4: Check if |M ∩ P ∩ C| = 20 is feasible.

Any triple intersection cannot exceed any pairwise intersection, because:

|M\cap P\cap C|\le |M\cap P|\le 11

So |M ∩ P ∩ C| cannot exceed 11. The value of 20 violates this constraint, so we cannot use all pairwise intersections at their maximum simultaneously.

Step 5: Re-examine with the binding constraint.

We need:

|M\cap P\cap C|\le |M\cap P|\le 11

So the absolute upper bound from the pairwise constraint is 11. Let us verify whether |M ∩ P ∩ C| = 11 is achievable by checking the inclusion-exclusion equation:

Let x = |M ∩ P ∩ C| = 11. We need the pairwise intersections to satisfy:

|M\cap P|+|P\cap C|+|M\cap C|=61-40+11=32

We need |M ∩ P| + |P ∩ C| + |M ∩ C| = 32, with each pairwise value ≥ x = 11 (since the triple is part of each pair) and within the given at-most bounds (11, 15, 15).

A possible assignment: |M ∩ P| = 11, |P ∩ C| = 11, |M ∩ C| = 10.
But |M ∩ C| ≥ |M ∩ P ∩ C| = 11 is required, so |M ∩ C| = 10 < 11 is invalid.

Try: |M ∩ P| = 11, |P ∩ C| = 11, |M ∩ C| = 10 — still invalid.
Try: |M ∩ P| = 11, |P ∩ C| = 12, |M ∩ C| = 9 — invalid (|M ∩ C| < 11).

Every pairwise intersection must be ≥ 11 (the triple). So minimum sum of pairwise = 11 + 11 + 11 = 33 > 32. This is a contradiction! So |M ∩ P ∩ C| = 11 is not achievable either.

Step 6: Find the true maximum using the correct bound.

Let x = |M ∩ P ∩ C|. Then each pairwise intersection ≥ x, so:

|M\cap P|+|P\cap C|+|M\cap C|\ge 3x

From Step 2:

x=-21+\left(\right|M\cap P|+|P\cap C|+|M\cap C\left|\right)

Since |M ∩ P| + |P ∩ C| + |M ∩ C| ≥ 3x:

x\ge -21+3x

21\ge 3x-x=2x

x\le \frac{21}{2}=10.5

Since x must be a whole number (count of students):

x\le 10

Step 7: Verify x = 10 is achievable.

Set |M ∩ P ∩ C| = 10. Then from the inclusion-exclusion equation:

|M\cap P|+|P\cap C|+|M\cap C|=61-40+10=31

We need values with each pairwise ≥ 10 and within the given bounds. Try:

|M ∩ P| = 11,   |P ∩ C| = 10,   |M ∩ C| = 10
Sum = 11 + 10 + 10 = 31 ✓
All ≥ 10 ✓
All within the at-most limits (11 ≤ 11, 10 ≤ 15, 10 ≤ 15) ✓

This is a valid configuration, confirming that x = 10 is achievable.

Conclusion: The maximum number of students who passed in all three subjects is 10.