To solve the problem, we use the principle of inclusion-exclusion for three sets and aim to maximize the number of students taking all three subjects.

Step 1: Define Variables

Let (P) (Physics), (C) (Chemistry), (M) (Math) be the sets of students taking each subject. Let (x = |P∩C∩M|) (students taking all three). Let (a = |P∩C| - x), (b = |P∩M| - x), (c = |C∩M| - x) (students taking exactly two subjects).

Since every student took at least one subject:
(|P∪C∪M| = 25).

Step 2: Apply Inclusion-Exclusion

The formula:
[|P∪C∪M| = |P| + |C| + |M| - |P∩C| - |P∩M| - |C∩M| + |P∩C∩M|]

Substitute values:
[25 = 12 + 15 + 18 - (a+x) - (b+x) - (c+x) + x]

Simplify:
[25 = 45 - (a+b+c + 2x)]
[a + b + c + 2x = 20]

Step 3: Maximize (x)

To maximize (x), minimize (a+b+c) (minimum is 0):
[2x ≤ 20 → x ≤ 10]

Check if (x=10) is valid:

• Only (P): (12 - 10 = 2)
• Only (C): (15 - 10 = 5)
• Only (M): (18 - 10 = 8)
• All three: 10

Total: (2+5+8+10=25) (matches the class size).

Answer: (\boxed{10})

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