SOLUTION: In a survey of 100 out patients who were reported at a hospital one day found out that 70 complained of fever,50 complained of stomach ache and 30 were injured. All 100 patients ha

Let F be the set and the number of those who complained of fever        (F = 70);
    S be the set and the number of those who complained of stomach ache (S = 50);
    I be the set and the number of those who wee injured                (I = 30).


Do not worry that I denoted by the same symbol the set and the number: I made it for simplicity, 
and you always can distinct from the context what I am talking about.


Let FS be the intersection of the sets F and S and the number of elements in this set at the same time.

Let FI be the intersection of the sets F and I and the number of elements in this set at the same time.

Let SI be the intersection of the sets S and I and the number of elements in this set at the same time.



    Let FSI = x be the intersection of the sets F, S and I and the number of elements in this intersection at the same time.

    I called the last quantity as "x", since it is our major unknown in this problem.


SO,        the set FS includes those who has two complains F and S, but also includes those who has all 3 complains. (*)
Similarly, the set FI includes those who has two complains F and I, but also includes those who has all 3 complains. (*)
Finally,   the set SI includes those who has two complains S and I, but also includes those who has all 3 complains. (*)

   
From the elementary theory of finite sets, we have this equation 

100 = F + S + I - FS - FI - SI + x.       (1)


From the given part of the condition, we have this equation

(FS-x) + (FI-x) + (SI-x) = 44.            (2)


    Indeed,    (FS-x) are those and only those who has exactly two complains F and S and do not have third complain.  (*)
    Similarly, (FI-x) are those and only those who has exactly two complains F and I and do not have third complain.  (*)
    Finally,   (SI-x) are those and only those who has exactly two complains S and I and do not have third complain.  (*)
    So, the left part of (2) are those and only those who have exactly two complains - exactly as its right side.     (*)


Now, we can re-write equation (1) in this form

100 = 70 + 50 + 30 - (FS-x) - (FI-x) - (SI-x) -3x + x = 

    = 150 - [(FS-x) + (FI-x) + (SI-x)] - 2x = 150 - 44 - 2x = 106 - 2x,

which gives us 

2x = 106 - 100 = 6.


Hence,  x = 3.


Answer.  The number of patients who had all three complaints was 3.