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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, and 20 had milk.
How many students had nothing?
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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
Solved.
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For Inclusion-Exclusion principle see this Wikipedia article
https://en.wikipedia.org/wiki/Inclusion%E2%80%93exclusion_principle
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