Arrangement of letters in VISITING with no pairs of consecutive Is. Could someone help me understand a book's solution to the following problem?  I am providing my own solution, but I fail to understand theirs.
Question:

How many ways are there to arrange the letters in VISITING with no
  pair of consecutive Is?

My Solution:


*

*There are three Is and five other letters.  I first arrange the five other letters with a space between each one, in the beginning and in the end.  There are $5!$ ways of arranging them:  _X_X_X_X_X_

*The previous step creates six empty places in which three Is may be arranged, but only three places must be chosen.  Order does not matter, as all three Is are the same.
Thus, there are $\dbinom{6}{3}$ ways of choosing those places.


My Answer: $5!\cdot\dbinom{6}{3}$
Book's Answer: $\dbinom{(5-2)+4-1}{(5-2)}\cdot5!$
Both answers yield the same numeric result of $2400$, but I don't fully understand their answer or rather how they achieved it.  I'm sure in fact that the second part, $5!$, is the arrangement of five other letters, other than Is.  This leaves the first part of the answer, which is selection with repetitions.  I do not know what $(5-2)$ and $4$ parts represent, especially $4$.
 A: A gentleman on ##math freenode IRC channel was kind enough to explain this to me, so here's the solution for anyone wondering:
Solution:


*

*Arrange three Is such that there is at least one place for other letters between each one: _I_I_I_

*Based on that arrangement, there are a total of four places where other letters may be placed, but there must be at least one letter in each of the two middle spots marked by X: _IXIXI _

*Since there were five letters other than I to begin with and two of them must be placed in those two marked spots, it leaves three letters to be distributed between the four positions around Is $\dbinom{3+4-1}{3}$

*Since the first step was picking the spots, we now must arrange the five letters between those spots and there are $5!$ ways of doing so.


Answer: $\dbinom{3+4−1}{3}\cdot5!$ or $\dbinom{(5-2)+4-1}{(5-2)}\cdot5!$
A: Here's one more solution, and solves the general kind of pattern avoidance.
Construct a matrix
\begin{align*}
 A &= \begin{array}{|l|rrrrrr|}\hline
 & \mathrm{V} & \mathrm{I} & \mathrm{S} & \mathrm{T} & \mathrm{N} & \mathrm{G} \\ \hline
\mathrm{V} & v & i & s & t & n & g \\
\mathrm{I} & v & 0 & s & t & n & g \\
\mathrm{S} & v & i & s & t & n & g \\
\mathrm{T} & v & i & s & t & n & g \\
\mathrm{N} & v & i & s & t & n & g \\
\mathrm{G} & v & i & s & t & n & g \\ \hline 
\end{array}
\end{align*}
The labels with caps are the unique symbols, and the small letters are the variables indicating the number of such symbols from which the pattern is to be formed.
Since $I$ can't be adjacent to $I$, that entry is made 0.
Now, the recurrence relation can be formed from the matrix by finding the characteristic polynomial:
\begin{align*}
  -g i x^4-g x^5-i n x^4-i s x^4-i t x^4-i v x^4-n x^5-s x^5-t x^5-v x^5+x^6
\end{align*}
and the recurrence is:
\begin{align*}
  f(v,i,s,t,n,g) &= f(v-1,i,s,t,n,g)+f(v,i,s-1,t,n,g)+f(v,i,s,t-1,n,g)+f(v,i,s,t,n-1,g)+f(v,i,s,t,n,g-1)+f(v-1,i-1,s,t,n,g)+f(v,i-1,s-1,t,n,g)+f(v,i-1,s,t-1,n,g)+f(v,i-1,s,t,n-1,g)+f(v,i-1,s,t,n,g-1)
\end{align*}
The following python code solves for the specific case where $i=3$ and the rest is 1
from math import factorial

def f(v,i,s,t,n,g):
    if v<0 or i<0 or s<0 or t<0 or n<0 or g<0:
        return 0
    if v<2 and i<2 and s<2 and t<2 and n<2 and g<2:
        return factorial(v+i+s+t+n+g)
    return f(v-1,i,s,t,n,g)+f(v,i,s-1,t,n,g)+f(v,i,s,t-1,n,g)+f(v,i,s,t,n-1,g)+f(v,i,s,t,n,g-1)+f(v-1,i-1,s,t,n,g)+f(v,i-1,s-1,t,n,g)+f(v,i-1,s,t-1,n,g)+f(v,i-1,s,t,n-1,g)+f(v,i-1,s,t,n,g-1)

print f(1,3,1,1,1,1) # = 2400

and also the g.f can be found by $\left(I- A\right)^{-1} = \dfrac{i+1}{1-(g (i+1)+i (n+s+t+v)+n+s+t+v)}$
