Question

A group of $123$ workers went to a canteen for cold drinks, ice-cream and tea, $42$ workers took ice-cream, $36$ tea and $30$ cold drinks. $15$ workers purchase ice cream and tea, $10$ ice cream and cold drinks, and $4$ cold drinks and tea but not ice cream, $11$ took ice cream and tea but not cold drinks. Determine how many workers did not purchase anything ?

Answer 1

Given $N=123$, $n\left(I\right)=42$, $n\left(T\right)=36$,
$n\left(C\right)=20$, $n(I\cap T)=15$, $n(I\cap C)=10$,
$n(C\cap T\cap I^{\prime })=4$, $n(I\cap T\cap C^{\prime })=11$,
To find $n(I^{\prime }\cap T^{\prime }\cap C^{\prime })$
We have
$4=n(C\cap T\cap I^{\prime })=n(C\cap T)-n(C\cap T\cap I^{\prime })$
and $11=n(I\cap T\cap C^{\prime })$
$=n(I\cap T)-n(I\cap T\cap C^{\prime })$
$=15-n(I\cap T\cap C)$
$\therefore n(I\cap T\cap C)=4$
Then $\left(1\right)$ Gives $4=n(C\cap T)-4$
or $n(C\cap T)=8$
Now $n(I^{\prime }\cap T^{\prime }\cup C^{\prime })=n(I\cup T\cup C{)}^{\prime }$
by De-Morgan law
$=N-n(I\cup T\cup C)$
$=N-\{n\left(I\right)+n\left(T\right)+n\left(C\right)-n(I\cap T)-n(T\cap C)-n(C\cap I)+n(I\cap T\cap C)\}$
$=123-\{42+36+30-15-8-10+4\}=44$