SOLUTION: Sixty-five students went to zoo with: 3 had hamburger, milk, and cake, 5 had hamburger and milk, 10 had cake and milk, 10 had hamburger and cake, 24 had hamburger, 38 had cake,

You are given an universal set U of 65 students.

You are given 3 its subsets

    H  of 24 students  (hamburger)

    C  of 38 students  (cake)

    M  of 20 students  (milk)


You are given their in-pairs intersections 

    HC  with 10 elements (it is the intersection H and C: hamburger and cake);

    HM  with  5 elements (it is the intersection H and M: hamburger and milk);

    CM  with 10 elements (it is the intersection C and M: cake and milk).


Also, you are given the set of their triple intersection HCM of 3 students (hamburger, cake and milk).



Now apply the REMARCABLE formula for the union of three subsets  (H U C U M)  in the universal set 

    n(H U C U M) = n(H) + n(C) + n(M) - n(HC) - n(HM) - n(CM) + n(HCM).


This formula is called  "the inclusion-exclusion principle".  It is WELL KNOWN in Elementary set theory.


Substitute the known values into the formula.

    n(H U C U M) = 24 + 38 + 20 - 10 - 5 - 10 + 3 = 60.


The set  (H U C U M)  is the set of students that have at least one food.


Hence, the rest  65 - 60 = 5 students do not have any food.   ANSWER