1st patient occupies any of 10 beds (10 options) 2nd patient occupies any of remaining 9 beds (9 options) 3rd patient occupies any of remaining 8 beds (8 options) 4th patient occupies any of remaining 7 beds (7 options) 5th patient occupies any of remaining 6 beds (6 options) THEREFORE, there are 10*9*8*7*6 = 30240 different ways to accommodate. ANSWER