smallest intersection of 4 subsets In a small town of 100 men, 85 are married, 70 have a telephone, 75 own a car, and 80 own their own home. On this basis, what is the smallest possible number of men who are married, have a telephone, own a car, and own their own home?
I tried the Venn diagram but it seems more complicated. What would be the right way to solve this? Thanks!
 A: Let $M$, $T$, $C$, and $H$ be the set of people who are married, have a telephone, own a car, and own their own house, respectively. Then, with $()^\complement$ denoting the complement
 set,
\begin{align}
 |M\cap T\cap C\cap H|&=100 - |(M\cap T\cap C\cap H)^\complement|\\
& = 100 - |M^\complement\cup T^\complement\cup C^\complement\cup H^\complement|\tag{1}\\
&\geq 100 - (|M^\complement|+|T^\complement|+| C^\complement|+|H^\complement|) \tag{2}\\
&= 100-(15+30+25+20)=10,\end{align}
where (1) follows from the De Morgan's law and (2) follows from the union bound.
A: The answer is $10$.  For each new property considered, place as many men outside the conjunction set as possible.  (Read the plot from bottom to top.)

A: $85$ men are married.  So $15$ are not.
$70$ men have telephones.  The very most of those that are unmarried is $15$.  So the very least that aren't married are $70-15=55$.  And the very most of the men that are not both married and telephone owners is $100 -55=45$.
$75$ men have cars.  At most $45$ of these aren't both married and telephone owners.  So at least $30$ are are married telephone owners.  So of all the $100$ men the very most that are not married telephone owning car owners is $70$.
$80$ men own there own home.  At the most $70$ are not married telephone owning car owners.  So at least $80-70=10$ are all four.
