Let A, B, and W denote the sets of students apply to Addis Ababa Uni (A), Bahir Dar Uni (B), or Wachemo Uni (W). Let U denote the universal set of all students in the class.
We're given the cardinalities of several sets:
• total number of students: [tex]n(U) = 60[/tex]
• A applicants: [tex]n(A) = 30[/tex]
• B applicants: [tex]n(B) = 25[/tex]
• W applicants: [tex]n(W) = 24[/tex]
• A and B applicants: [tex]n(A\cap B) = 11[/tex]
• A and W applicants: [tex]n(A \cap W) = 6[/tex]
• B and W applicants: [tex]n(B\cap W) = 9[/tex]
• non-applicants: [tex]n(U \setminus (A\cup B\cup W)) = 4[/tex]
The last cardinality tells us [tex]n(A\cup B\cup W) = 60-4 = 56[/tex] students applied anywhere at all.
We want to find [tex]n(A\cap B\cap W)[/tex], the number of students that applied to each of the three universities.
By the inclusion/exclusion principle,
[tex]n(A\cup B\cup W) = n(A) + n(B) + n(W) \\\\~~~~~~~~~~~~~~~~~~~~~~~~ - n(A\cap B) - n(A\cap W) - n(B\cap W)\\ \\~~~~~~~~~~~~~~~~~~~~~~~~ + n(A\cap B\cap W)[/tex]
That is, we count up all the students in the sets A, B, and W, then subtract the number of students in each pairwise intersection to not double-count, then add back the number of students in the intersection of all three sets since it was removed in the previous step.
Now solve.
[tex]56 = 30 + 25 + 24 - 11 - 6 - 9 + n(A\cap B\cap W)[/tex]
[tex]\implies n(A\cap B\cap W) = \boxed{3}[/tex]
