1. The problem statement, all variables and given/known data Decide in how many ways the letters of the word ABRACADABRA can be arranged in a row if C, R and D are not to be together. 2. Relevant equations The number of ways of arranging n objects which include 'a' identical objects of one type, 'b' identical objects of another type,.... is n!/(a!b!....) n objects divided into m groups with each group having G1, G2, ..., Gm objects respectively has m! * G1! * G2! * ... *Gm! 3. The attempt at a solution A: 5 B: 2 R: 2 C: 1 D: 1 11 letters in total. There are a few identical letters so the total number of ways of permuting the objects accounting for the identical letters is 11!/(5!2!2!) The total number of ways of arranging the letters such that C, R and D are together is: 8!4!/(5!2!2!) as there are 8 groups, with one group containing 4 letters. However given the identical letters, we divide by the same number as above. So (11!-8!4!)/(5!2!2!)=81144 The answers suggested 78624 I can't see what is wrong with my reasoning.