reto wrote:
How many different arrangements are possible to place seven different books on a shelf if all three math books must be placed next to each other?
A. 120
B. 148
C. 360
D. 540
E. 720
Let the 3 math books be \(M_1, M_2, M_3\) while the rest be A,B,C,D for a total of 7
different books.
Consider 3 math books to be 1 collective item (=\(M_1M_2M_3\) = M , as they need to be together). Thus, number of arrangements of 3 different books (\(M_1M_2M_3\))= 3P3 = 3!
Total number of arrangements of M,A,B,C,D = 5P5= 5!
Thus, The total possible arrangements will be : \(M_1M_2M_3ABCD\) = 5!*3! = 720, E is the correct answer.