In a class of 400 students, 180 read English, 371 read Sanskrit, 270 read Hindi, at least how many students read all the three?
No other information is given about how many students reading any two of the given subjects. I want to use $\#(A\cup B \cup C)$ formula but I don't have sufficient information about $\#(A\cap B)$ or $\#(B \cap C)$ or $\#(A\cap C).$ I don't have other ideas for dealing with this problem. How can I solve this problem?
$400 - 371 = 29$ students do not read sanskrit.
$270$ students read Hindi. At most $29$ of them do not read Sanskrit. So at least $270 - 29 = 241$ students read both Hindi and Sanskrit.
So at most $400 - 241 = 159$ students don't read both.
$180$ read english and at most $159$ of the don't read both sanskrit and hindi. So at least $180 - 159 = 21$ of the students who read english read both sanskrit and hindi.
So at the very least $21$ students read all three.
....
To do it by exclusion/inclusion.
$|E \cup S \cup H| = 400$
$|E \cup S \cup H|= |E|+ |S| + |H| - (|E\cap S| + |E \cap H| + |S \cap H|) + |E \cap S \cap H|=180 + 371 + 270 - (|E\cap S| + |E \cap H| + |S \cap H|) + |E \cap S \cap H|$
So $|E \cap S \cap H| = (|E\cup S| + |E \cup H| + |S \cup H| )- 421$.
Now $400 \ge |E\cup H| = |E| + |H| - |E\cap H|= 180+270- |E\cap H|$ so $|E\cap H|\ge 50$.
$400\ge |E\cap S| = 180 + 371 + |E\cap S|$ so $|E\cap S| \ge 151$
$400\ge |H\cap S| = 270 + 371 + |H\cap S|$ so $|H\cap S| \ge 241$
So $|E \cup S \cup H| = (|E\cap S| + |E \cap H| + |S \cap H| )- 421\ge 50 + 151 + 241 - 421 = 21$