In how many ways can the word "INDIVISIBILITY" be rearranged, such
that no two Is come close to each other?
My attempt
Total number of ways of rearranging the word = $\dfrac{14!}{6!}$
Number of ways of rearranging the word such that at the least, a pair of Is come close to each other (the motivation for this comes from the negation of "no two Is come close to each other") = $\binom{6}{2} \times \dfrac{13!}{4!}$
Then, required number of arrangements = $\dfrac{14!}{6!} – \binom{6}{2} \times \dfrac{13!}{4!}$.
But this gives me a negative number. Where am I going wrong? Could anyone tell me how to go about this problem?
Best Answer
There are $8$ distinct letters, together with $6$ I's.
The $8$ non-I's can be arranged in $8!$ ways. Every such arrangement produces $9$ 'gaps" (the $7$ gaps between letters, and the $2$ 'endgaps") for us to slip a single I into. The $6$ gaps needed can be chosen in $\binom{9}{6}$ ways, for a total of $8!\binom{9}{6}$.
Remark: The $\binom{6}{2}\frac{13!}{4!}$ overcounts the cases where two or more I's are next to each other.