(c)
Given, $n(M) = 100,\ n(P) = 70,\ \ \ \ \ n(C) = 40$
$\ \ \ \ \ \ \ \ n(M \cap P) = 30,\ \ \ \ \ n(M \cap C) = 28,$
$\ \ \ \ \ \ \ \ \ n(P \cap C) = 23$ and $n(M \cap P \cap C) = 18$
$\therefore n\left( M \cap P^{'} \cap C^{'} \right) = n\lbrack M \cap (P \cap C')\rbrack$
$= n(M) - n\left\lbrack M \cap (P \cap C) \right\rbrack$
$= n(M) - \lbrack n(M \cap P) + n(M \cap C) - n(M \cap P \cap C)\rbrack$
$= 100 - \lbrack 30 + 28 - 18 = 60\rbrack$