Out of 800 boys in a school, 224 played cricket, 240 played hockey and 336 played basketball. Of the total, 64 played both basketball and hockey; 80 played cricket and basketball and 40 played cricket and hockey, 24 played all the three games. The number of boys who did not play any game is

Given $n\left(C\right)=224,n\left(H\right)=240,n\left(B\right)=336,n(H\cap B)=64,n(B\cap C)=80,n(H\cap C)=40$, $n(C\cap H\cap B)=24$

We know that $n(C^C\cap H^C\cap B^C)=n\left[{(C\cup H\cup B)}^C\right]=n\left(U\right)-n(C\cup H\cup B)$

Substitute the appropriate values in the above equation

$\begin{array}{rcl}n(C^C\cap H^C\cap B^C)& =& 800-\left[n\left(C\right)+n\left(H\right)+n\left(B\right)-n(H\cap C)-n(H\cap B)-n(C\cap B)+n(C\cap H\cap B)\right]\\ & =& 800-[224+240+336-64-80-40+24]\\ & =& 800-[824-184]\\ & =& 984-824\\ & =& 160\end{array}$

Therefore, $\begin{array}{rcl}n(C^C\cap H^C\cap B^C)& =& \boxed{160}\end{array}$