It should be four big and three small.
There are X cars and Y cars.
There are 2X+Y= 1 1 (number of doctors), Y= 1 1-2X.
Since the last car is required to have at least 20 seats, the seats in the last car will not exceed 10, and the seats in all cars will not exceed 253+ 1 1 (doctor)+10 = 274.
274 >45X+30Y & gt; 264, substituting Y= 1 1-2X,
274 >330- 15X & gt; 264, 56