We have a total of 8 letters. Some of the letters appear more than once. The requirement of the question stem that the letter C is to be to the right of the letter D, with the letters D and C each appearing once.
If n is a positive integer, the number of different ways to arrange n different objects is n!, where n! is the product of the first n positive integers. Thus, n! = n (n - 1)(n - 2) … (3)(2)(1).
If we had 8 different letters, the number different of ways to arrange the 8 letters would be 8! = (8)(7)(6)(5)(4)(3)(2)(1). However, this number will be lower because of duplicates in this problem. Because the letter A appears twice, we must divide 8! by 2! Because the letter B appears 3 times, we must divide 8! by 3!. So the number of different ways to arrange the letters A, A, B, B, B, C, D, E is . We also have the requirement that the letter C is to the right of the letter D. For any specified locations of the letter C and the letter D, there are the same number of arrangements for the other 6 letters when C is to the right of D and D is to the right of C. So we must divide the number of possible arrangements of all the letters by 2. So the number of different ways to arrange the letters A, A, B, B, B, C, D, E where the letter D is to the right of the letter C is , which is . To find the answer to the question, let's find the value of . We have = = = (4)(7)(3)(5)(4) = 1,680. Answer Choice (A) is correct.
