SOLUTION: Find the number of ways 5 large books, 4 medium-size books, and 3 small books can be placed on a shelf where: (a) there are no restrictions; (b) all books of the same size are tog

    The total number of books is 5 + 4 + 3 = 12.


    The total number of all possible arrangements is equal to the number of all permutations of 12 objects 


         12! = 12*11*10*9*8*7*6*5*4*3*2*1 = 479001600.      ANSWER

     First, we consider the block of 5 large books, the block of 4 medium-size books and the block of 3 small books 

     as 3 separate entities. These entities can be arranged in 3! = 6 ways as the blocks.


     Now, 5! = 120 permutations are possible inside the block of 5 large books;

          4! =  24 permutations are possible inside the block of 4 medium-size books, and

          3! = 6 permutations are possible inside the block of 3 small books.


     All these permutations are INDEPENDENT; therefore, the total number arrangements of the books on the shelf 
     under the given restriction is the product


              6*120*24*6 = 103680 ways.      ANSWER