We are given that 45 students are participating in exams, and each student must take at least one of the three exams: Mathematics, Physics, and Chemistry. We also have the following information:

• 25 students take the Mathematics exam.
• 15 students take the Physics exam.
• 19 students take the Chemistry exam.
• 9 students take at least two exams.

We need to find how many students take all three exams.

Let:

• $n(M) = 25$ be the number of students taking the Mathematics exam,
• $n(P) = 15$ be the number of students taking the Physics exam,
• $n(C) = 19$ be the number of students taking the Chemistry exam,
• $n(M \cap P \cap C) = x$ be the number of students taking all three exams (this is what we need to find),
• $n(M \cup P \cup C) = 45$ be the total number of students (since every student takes at least one exam).

We are also told that 9 students take at least two exams. The number of students taking exactly two exams can be expressed as:

$n(M \cap P) + n(P \cap C) + n(C \cap M) - 3x = 9$

Using the inclusion-exclusion principle, the total number of students taking at least one exam is:

$n(M \cup P \cup C) = n(M) + n(P) + n(C) - [n(M \cap P) + n(P \cap C) + n(C \cap M)] + n(M \cap P \cap C)$

Substitute the values we know:

$45 = 25 + 15 + 19 - [n(M \cap P) + n(P \cap C) + n(C \cap M)] + x$

Simplify:

$45 = 59 - [n(M \cap P) + n(P \cap C) + n(C \cap M)] + x$

From this, we get:

$n(M \cap P) + n(P \cap C) + n(C \cap M) = 14 + x$

Now, substitute this into the equation for students taking at least two exams:

$14 + x - 3x = 9$

Simplify:

$14 - 2x = 9$

Solve for $x$:

$14 - 9 = 2x$ $x = \frac{5}{2} = 2.5$ Step use