Let $X$ denote the cars having CD, $Y$ those having AC and $Z$ those having white wall tyres.
$\mid X\mid = 20$
$\mid Y\mid = 10$
$\mid Z \mid = 8$
No. of cars having at least one option is given by
$S = \mid X \cup Y \cup Z \mid = \mid X \mid + \mid Y\mid + \mid Z \mid - \mid X \cap Y \mid - \mid X \cap Z \mid - \mid Y \cap Z \mid + \mid X \cap Y \cap Z \mid.$
No. of cars which do not have any option at all, $M$ = Total no. of cars - No. of cars having at least one option.
$= 40 - S.$
We are asked to find the minimum number of cars having no option. So, we have to minimize $M$ which corresponds to maximizing $S$, which again corresponds to minimizing
$\mid X \cap Y,$
$\mid X \cap Z$ and
$\mid Y \cap Z$.
We have $\mid X \cap Y \cap Z \mid = 5$
So,
$\mid X \cap Y \geq 5.$
$\mid X \cap Z \geq 5.$
$\mid Y \cap Z \geq 5.$
Thus we get the minimal values of all three as $5$ which gives
$S = 20 + 10 + 8 - 5 - 5-5 + 5 = 28 \\ \implies M = 40 - 28 = 12.$
Suppose we are asked to find the maximum possible cars which can be without any option, we have to maximize $M$ and so minimize $S$.
From set intersection property we can also write
$\mid X \cap Y \leq \min\left(\mid X \mid, \mid Y \mid\right).$
which gives
$\mid X \cap Y \mid \leq \min(20, 10) \leq 10$.
Similarly
$\mid X \cap Z \mid \leq \min(20, 8) \leq 8$ and
$\mid Y \cap Z \mid \leq \min(10, 8) \leq 8$.
So, if we maximize these three, we get
$S = 20 + 10 + 8 - 10 - 8 - 8 + 5 = 17.$ and
$M = 40 -17 = 23.$
So, at most 23 cars can be there without any option.
