In how many ways can one arrange all of the letters in the word INFORMATION so that no pair of consecutive letters appears more than once. Eg ININformota is not acceptable as IN appears twice.
So in know there are $11!$ different arrangements of the word information. I need to use the inclusion and exclusion principle to eliminate the words that have repeated pairs of words. I am looking for a hint on how to go about counting the total number of words where IN is repeated twice and then NI is repeated twice, similarly where ON and NO are repeated, IO and OI are repeated. I guess hint whether ION and NOI would have to be considered or not.
Best Answer
So this is my own attempt at the question.
So total possible ways to permute INFORMATION is $N = 11!/{2!^3} = S_0$
Let $c_1$ be the total number of permutations in which IN and NI is repeated twice $=2*9!/{2!^2}$ $c_2$ is where ON and NO is repeated and $c_3$ is where $IO$ and OI is repeated and $c_1=c_2=c_3$ therefore $S_1 = 3*c_1$
now there are also the three letter words ION and NOI that are little tricky and there are $3!$ ways they can be arranged and thus they are are $3!*7!/{2!}$ lets call this $S_2$
so I am thinkinking it should be $S_0-S_1+S_2 = 4989600-544320+15120=4460400$