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

We have 40 students who passed at least one of three subjects M, P, and C, with |M| = 20, |P| = 25, |C| = 16, |M ∩ P| ≤ 11, |P ∩ C| ≤ 15, and |M ∩ C| ≤ 15. We wish to determine the maximum number of students who passed all three subjects.

Since every student passed at least one subject, the Inclusion-Exclusion Principle gives

$$|M ∪ P ∪ C| = |M| + |P| + |C| - |M ∩ P| - |P ∩ C| - |M ∩ C| + |M ∩ P ∩ C| = 40$$

Substituting the known totals,

$$20 + 25 + 16 - |M ∩ P| - |P ∩ C| - |M ∩ C| + |M ∩ P ∩ C| = 40$$

which simplifies to

$$61 - \bigl(|M ∩ P| + |P ∩ C| + |M ∩ C|\bigr) + |M ∩ P ∩ C| = 40\,. $$

Letting $$S = |M ∩ P| + |P ∩ C| + |M ∩ C|$$ and $$t = |M ∩ P ∩ C|$$, this equation becomes

$$61 - S + t = 40\quad\Longrightarrow\quad S - t = 21\,,\quad\text{so}\quad S = 21 + t\,. $$

Each pairwise intersection is at most 11, 15, and 15 respectively, and must be at least $$t$$ since the triple intersection is contained in each pair. Hence the maximum possible sum is $$S=11+15+15=41$$, giving $$t ≤ 41 - 21 = 20$$, while also $$t ≤ |M ∩ P| ≤ 11$$. Thus $$t ≤ 11$$.

We must also ensure that every region in the Venn diagram has a non-negative number of students. For $$t = 10$$ we have $$S = 31$$ and may choose $$|M ∩ P| = 11\,,\; |P ∩ C| = 10\,,\; |M ∩ C| = 10\,. $$ Then

Only-M = $$20 - 11 - 10 + 10 = 9 \geq 0$$ $$\checkmark$$; Only-P = $$25 - 11 - 10 + 10 = 14 \geq 0$$ $$\checkmark$$; Only-C = $$16 - 10 - 10 + 10 = 6 \geq 0$$ $$\checkmark$$; (M ∩ P) only = $$11 - 10 = 1 \geq 0$$ $$\checkmark$$; (P ∩ C) only = $$10 - 10 = 0 \geq 0$$ $$\checkmark$$; (M ∩ C) only = $$10 - 10 = 0 \geq 0$$ $$\checkmark$$; and the total is $$9 + 14 + 6 + 1 + 0 + 0 + 10 = 40$$ $$\checkmark$$.

For $$t = 11$$ we would have $$S = 32$$. One might take $$|M ∩ P| = 11$$, $$|P ∩ C| = 11$$, $$|M ∩ C| = 10$$, giving Only-M = $$20 - 11 - 10 + 11 = 10 \geq 0$$, Only-P = $$25 - 11 - 11 + 11 = 14 \geq 0$$, Only-C = $$16 - 11 - 10 + 11 = 6 \geq 0$$, but then (M ∩ C) only = $$10 - 11 = -1 < 0$$ $$\times$$. Any other choice with $$|M ∩ P|=11$$ requires the other two intersections to sum to 21 while each is at least 11, forcing a sum of at least 22, which is impossible. Hence $$t = 11$$ cannot be realized.

These considerations show that the maximum feasible value of $$|M ∩ P ∩ C|$$ is 10.

The answer is 10.