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,

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Question 1178801: 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?
Answer by ikleyn(52858)   (Show Source): You can put this solution on YOUR website!
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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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