55 people are enrolled in courses of the Diploma of Information Technology. While 35 are registered in OOP, 23 in BD, 27 in SAP, 15 in OOP and SAP, 9 in SAP and BD, 12 in OOP courses and BD and, finally, 4 are enrolled in the three courses, how many people are not enrolled in any of these three courses?
My attempt: $55−(((35−(15+12))+(27−(15+9))+(23−(12+9)))+(12+15+9))=2$
Your answer is correct.
Let $O$ denote the set of students enrolled in $OOP$; let $B$ denote the number of students enrolled in $BD$; let $S$ denote the number of students enrolled in $SAP$. The Inclusion-Exclusion Principle states that the number of students taking at least one of the three courses is $$|O \cup B \cup S| = |O| + |B| + |S| - |O \cap B| - |O \cap S| - |B \cap S| + |O \cap B \cap S|$$ Hence, the number of students enrolled in at least one of the courses is $$|O \cup B \cup S| = 35 + 23 + 27 - 12 - 15 - 9 + 4 = 53$$ Therefore, the number of students enrolled in none of these courses is $$|O^C \cap B^C \cap S^C| = |U| - |O \cup B \cup C| = 55 - 53 = 2$$