In a survey of the 100 out-patients who reported at a hospital one day, it was found out that 70 complained of fever, 50 complained of stomach ache and 30 were injured. All 100 patients had at least one of the complaints and 44 had exactly two of the complaints. How many patients had all three complaints?
Let $a$ of them complaining just about the fever, $b$ of stomach and $c$ were injured. Also lets consider that $d$ of them were complaining about the fever and stomach, $e$ about stomach and injury, and $f$ about the injury and fever. Finally, let $g$ the number of them who have all three complaints. We have $a+b+c+d+e+f+g=100$, $a+d+f+g=70$, $b+d+e+g=50$, $c+e+f+g=30$. By adding the last 3 relations we obtain $a+b+c+2(d+e+f)+3g=150$, so $d+e+f+2g=50$. We also know that $d+e+f=44$, so $g=3$.
Hope I didn't mess up the calculus.