Anagrams and related problems I have a word like CONSTITUTIONALIST that is very fun for Anagram problems. So, in order to count the anagrams I have to: \begin{align*} s=\left\\{C(2),O(2),N(2),S(2),T(3),I(3),A(1)\right\\}\\\ P_{17!}^{2,2,2,2,3,3,1} = \dfrac{17!}{2!2!2!2!3!3!2!1!} \end{align*} Is that correct? So, if I wish to find out how many anagrams exists that have have no consecutive vowels repeated like, AA,EE...What would be the better way o solve? And if I would like to find out how many anagrams the I (that appears 3 times) never appear together 2 times ("II" may not occur). What would be the simplest way to solve?

**Hint** Let $a$ be the total amount of anagrams. Now if we want to count $a_{OO},$ the total amount of anagrams that have two consecutive "O", we may consider "OO" as a "new letter" and recount the total amount of anagrams (with one letter less and one different letter). Then there are exactly $a-a_{OO}$ anagrams that don't contains two consecutive $O$. Now, if we want to consider the number of anagrams that do not contain two consecutive "I", it is a little bit more tricky because there are three "I" to place. If we create a new letter "II" and count the number of anagram with it, then we will also count anagrams containing "II""I", so that $a-a_{II}$ is not equal to number of anagram without two consecutive "I". This is why you also have to count the number $a_{III}$ of anagrams with "III", so that the total number of anagrams without two consecutive "I" is given by $a-a_{II}-a_{III}$. Etc.