Combinatorics on the word Abracadabra How many different 'words' can be created using all the characters of '$ABRACADABRA'$? In how many of the 'words' that there are no identical characters one next to the other?
So, For the first part, the calculation would be $$\frac{11!}{5! 2! 2!} $$since there are 11 letters of which the letters $A$, $B$ and $R$ are repetitive.
But I can't quite figure out the second part of the question. Please help clarify it to me.
 A: Here is a different approach.  
According to Theorem 2.1 in Counting words with Laguerre series by Jair  Taylor, The Electronic Journal of Combinatorics 21(2) (2014) (here is a version of the paper at arXiv.org), the number of acceptable permutations is
$$\int_0^{\infty} e^{-t} \; \ell_1(t)^2 \; \ell_2(t)^2 \; \ell_5(t) \; dt$$
where
$$\begin{align}
\ell_1(t) &= t \\
\ell_2(t) &= \frac{1}{2} t^2 - t \\
\ell_5(t) &= \frac{1}{120} t^5 - \frac{1}{6} t^4 + t^3 - 2 t^2 + t \\
\end{align}$$
Mathematica evaluates the integral as $3084$.
A: We approach via inclusion-exclusion using the events:  no $A$'s are adjacent, no $B$'s are adjacent, no $R$'s are adjacent, labeled as events $\chi_a,\chi_b,\chi_r$ respectively.  Let $S$ be the universal event, the set of ways of arranging letters in abracadabra.
We are trying to count $|\chi_a\cap \chi_b\cap \chi_r|$
$|\chi_a\cap \chi_b\cap \chi_r| = |S\setminus(\chi_a^c\cup \chi_b^c\cup \chi_r^c)| = |S|-|\chi_a^c\cup \chi_b^c\cup \chi_r^c|$
$=|S|-|\chi_a^c|-|\chi_b^c|-|\chi_r^c|+|\chi_a^c\cap \chi_b^c|+|\chi_a^c\cap \chi_r^c| + |\chi_b^c\cap \chi_r^c| - |\chi_a^c\cap \chi_b^c\cap \chi_r^c|$
From earlier work, you already found $|S|=\binom{11}{5,2,2,1,1}=\frac{11!}{5!\cdot 2!\cdot 2!}$ correctly.
We ask ourselves, how do we calculate each of the remaining terms?  For $|\chi_a^c|$, the question is how many arrangements of abracadabra have at least two $A$'s adjacent.  The $A$'s present the most difficult problem to us in each of these terms since there are so many of them, so to continue, let us ask instead how many don't.  We know afterall that $|\chi_a|+|\chi_a^c|=|S|$.
In counting $|\chi_a|$, let us arrange the not-$A$ letters.  I.e. arrange the letters in the word brcdbr.  There are $\binom{6}{2,2,1,1}=\frac{6!}{2!\cdot 2!}$ such arrangements (using similar techniques as before).  Now that we have those letters arranged, give a bit of extra space to each side of every letter to make room for the $A$'s if one happens to be placed there.  From the $7$ available spaces (remember to count to the far left and to the far right as well) pick five of them to be occupied by the $A$'s.  In doing so, we guarantee that the $A$'s are not adjacent.  There are $\binom{7}{5}$ ways to complete this step.  Conclude then using multiplication principle that there are $\binom{6}{2,2,1,1}\binom{7}{5}$ to complete this process.  Convince yourself that this process counts every arrangement where no $A$'s are adjacent exactly once.  Thus, $|\chi_a|=\binom{6}{2,2,1,1}\binom{7}{5}$ and $|\chi_a^c| = \binom{11}{5,2,2,1,1}-\binom{6}{2,2,1,1}\binom{7}{5}$
Counting $|\chi_b^c|$ is easier since $|\chi_b^c|$ is the set of arrangements where the $B$'s are adjacent.  To count this, simply glue the two $b$'s together, perhaps like this: $~^b\!b$, and treat them like they were a single letter.  Via similar methods to the original problem, there are $\binom{10}{5,2,1,1,1}=\frac{10!}{5!\cdot 2!}$ arrangements.  Counting $|\chi_r^c|$ is the same.
Counting $|\chi_a^c\cap \chi_b^c|$ becomes a challenge again, but the method is the same as counting $|\chi_a^c|$.  We recognize that $|\chi_a^c\cap \chi_b^c| + |\chi_a\cap \chi_b^c|=|\chi_b^c|$, and counting $|\chi_a\cap \chi_b^c|$ is more manageable.  We use the same trick as before by gluing the $b$'s together, then arranging the not-$A$ letters, and then picking spaces for the $A$'s from those available, arriving at $|\chi_a\cap \chi_b^c| = \binom{5}{2,1,1,1}\binom{6}{5}$ implying $|\chi_a^c\cap \chi_b^c| = |\chi_b^c|-|\chi_a^c\cap \chi_b^c| = \binom{10}{5,2,1,1,1}-\binom{5}{2,1,1,1}\binom{6}{5}$.  Counting $|\chi_a^c\cap \chi_r^c|$ is identical.
Counting $|\chi_b^c\cap \chi_r^c|$ we use a similar technique as the others by gluing the $b$'s together and gluing the $r$'s together and arranging.
Finally, counting $|\chi_a^c\cap \chi_b^c\cap \chi_r^c|$ we recognize that $|\chi_a^c\cap \chi_b^c\cap \chi_r^c|+|\chi_a\cap \chi_b^c\cap \chi_r^c| = |\chi_b^c\cap \chi_r^c|$.  We use the exact same method as the previous times involving $A$'s, this time having to glue the $b$'s together and the $r$'s together before arranging everything.
Completing all necessary arithmetic will present the answer.

Having gone through the effort and seeing all of the cancellations going on, I recognize a slightly more efficient way: (calculation of the terms is same as above)
$|\chi_a^c\cup \chi_b^c\cup \chi_r^c| = |\chi_a^c|+|\chi_a\cap \chi_b^c| + |\chi_a\cap \chi_r^c| -|\chi_a\cap \chi_b^c\cap \chi_r^c|$  (this can be seen via a venn-diagram argument)
$=|S|-|\chi_a| +2|\chi_a\cap \chi_b^c| -|\chi_a\cap \chi_b^c\cap \chi_r^c|$ (using the symmetry of the cases of $b$'s and $r$'s)
So, $|\chi_a\cap \chi_b\cap \chi_r| = |\chi_a| - 2|\chi_a\cap \chi_b^c| + |\chi_a\cap \chi_b^c\cap \chi_r^c|$

 $=21\cdot 6!/4 - 2\cdot 5!/2\cdot 6 + 4! = 3084$

A: A Python 3 program that generates the permutations and counts the valid ones. I debugged it until it printed the value calculated by @JMoravitz. So I hope it now contains only negligible bugs.
s='abracadabra'
s=sorted(s)  #converts the string to a list and sorts it
count=0 # number of valid permutations
perm=0  # number of all permutation
increased=True
while increased:
    perm+=1
    # if this is a valid permutation 
    # ("no identical characters one next to the other")
    # then count it
    if increased: 
        valid=True
        for i in range(0,len(s)-1):
            if s[i]==s[i+1]:
                valid=False
                break
        if valid:
            count+=1
    # from the right find the first position where 
    # its right neighbor has a higher value
    # replace the substring (from this position to the end)
    # by the substring made of the same characters but following 
    # in lexical order the current substring
    increased=False
    for i in range(len(s)-1,0,-1):
        if s[i]>s[i-1]:
            maxchar=s[i]
            idxmax=i
            for j in range(i+1,len(s)):
                if s[j]>s[i-1]:
                    if s[j]<maxchar:
                        maxchar=s[j]
                        idxmax=j
            # swap the values and sort the remaining list
            s[idxmax]=s[i-1]
            s[i-1]=maxchar
            s[i:]=sorted(s[i:])
            increased=True
            break
print(count, perm)

Its output is
3084 83160

So there are 83160 different permutations and 3084 of them satisfy the constraint.
