How many ways are there to arrange the letters in INTELLIGENT with at least two consecutive pairs of identical letters?
I got an answer of
$$C(5,2)\cdot\frac{9!}{2!^3}-C(5,3)\cdot\frac{8!}{2!^2}+C(5,4)\cdot\frac{7!}{2!}-6!\;,$$
but the solution manual says
$$C(5,2)\cdot\frac{9!}{2!^3}-2\cdot C(5,3)\cdot\frac{8!}{2!^2}+3\cdot C(5,4)\cdot\frac{7!}{2!}-4\cdot 6!\;.$$
Where am I going wrong? can someone explain where the $2\cdot,3\cdot$, and $4\cdot$ are coming from in the solution manual?
Given subsets $A_1,\cdots, A_n$ of a finite set S,
let $T_k=\sum|A_{i_1}\cap\cdots\cap A_{i_k}|$ for $1\le k\le n$, where the sum is taken over all k-subsets of $\{1,\cdots n\}$.
If we let $N_m$ be the number of elements in at least m of the sets, then
$\displaystyle N_m=T_m-\binom{m}{m-1}T_{m+1}+\binom{m+1}{m-1}T_{m+2}-\binom{m+2}{m-1}T_{m+3}+\cdots+(-1)^{n-m}\binom{n-1}{m-1}T_n$,
as can be seen by determining how many times an element in r of the subsets is counted.
In this particular problem, $n=5$ and $m=2$.