A - radio

B - AC

C - power steering

Let n(cars that do not have any options at all)=x

n(A U B U C) + x = 50

so, x = 50 - n(A U B U C)

min(x) = 50 - max(n(A U B U C))

So what is max(n(A U B U C))?

n(A U B U C) = n(A) + n(B) + n(c) - n(AB) - n(BC) - n(CA) + n(ABC)

So all we need is min of each of n(AB), n(BC), n(CA) - So what is it?