Number of ways in which letters of the word ENGINEER can be arranged so that no two alike letters are together Number of ways in which letters of the word ENGINEER can be arranged so that no two alike letters are together is ________
My solution is as follows:
As per the figure number Es=3;G=1;I=1;Ns=2 and R=1, total word is 8

Number of arrangement of the word "ENGINEER" is $3360$ where $\frac{{8!}}{{3!.2!}} = 3360$
Now let's take the case that our focus is on E and neglect N initially.
Remove Case of the type EE,E,G,I,N,N,R is $\frac{{7!}}{{2!}} = 2520$
Now add cases when EEE,G,I,N,N,R is $\frac{{6!}}{{2!}} = 360$
Hence the number of ways Es are segregated is $3360-2520+360=1200$. There are number of cases when E are segregated. Now do we segregate N also?
 A: We can solve this problem with the Inclusion-Exclusion Principle.
The number of distinguishable arrangements of the letters of the word ENGINEER is
$$\binom{8}{3}\binom{5}{2}3!$$
since we can choose three of the eight positions for the Es, two of the remaining five positions for the Ns, then arrange the three distinct letters G, I, R in the remaining three positions.  From these arrangements, we must subtract those in which a pair of identical letters are adjacent.
A pair of adjacent identical letters:  There are two possibilities, a pair of Es is adjacent or a pair of Ns is adjacent.
A pair of Es is adjacent:  We have seven objects to arrange: EE, E, N, N, G, I, R.  Choose two of the seven positions for the Ns, then arrange the five distinct objects EE, E, G, I, R in the remaining positions, which can be done in $$\binom{7}{2}5!$$ ways.
A pair of Ns is adjacent:  We have seven objects to arrange:  E, E, E, NN, G, I, R.  Choose three of the seven positions for the Es, then arrange the four distinct objects NN, G, I, R in the remaining four positions, which can be done in $$\binom{7}{3}4!$$ ways.
Two pairs of adjacent identical letters:  There are again two possibilities, either there are two pairs of adjacent Es, in which case all three Es are consecutive, or there is a pair of adjacent Es and a pair of adjacent Ns.
Two pairs of adjacent Es:  We have six objects to arrange: EEE, N, N, G, I, R.  Choose two of the six positions for the Ns, then arrange the four distinct objects EEE, G, I, R in the remaining four positions, which can be done in $$\binom{6}{2}4!$$ ways.
A pair of adjacent Es and a pair of adjacent Ns:  We have six objects to arrange: EE, E, NN, G, I, R.  Since the six objects are distinct, they can be arranged in $$6!$$ ways.
Three pairs of adjacent identical letters:  The only way for this to occur is if there are two pairs of adjacent Es and a pair of adjacent Ns.  We have five objects to arrange:  EEE, NN, G, I, R.  Since the five objects are distinct, they can be arranged in $5!$ ways.
By the Inclusion-Exclusion Principle, the number of distinguishable arrangements of the letters of the word ENGINEER in which no two adjacent letters are identical is
$$\binom{8}{3}\binom{5}{2}3! - \binom{7}{2}5! - \binom{7}{3}4! + \binom{6}{2}4! + 6! - 5!$$
A: Arrangements with $E's$ placed apart in gaps$\;(-)\;$ [$N's$ may be together or separate]
eg $-N-G-I-N-R-\;\; :\;\; \binom63\times \frac{5!}{2!} = 1200$
Subtract arrangements with the two $N's$ bunched  as a super N:$\;\binom53\times 4! = 240$
Answer: $1200-240 = 960$
A: To construct the words where no two same characters adjacent ,we use Smirnov words
A generating function for the number of Smirnov words over an $n$-ary alphabet is given by
\begin{align*}
\left(1-\frac{nz}{1+z}\right)^{-1}\
\end{align*}

Here we consider an alphabet $\mathcal{V}=\{E,G,I,N,R\}$ with $n=5$ letters. Using $[z^k]$ to denote the coefficient of $z^k$ of a series we calculate
\begin{align*}
&\color{green}{[E^3GIRN^2]\left(1-\sum_{a\in\mathcal{V}}\frac{a}{1+a}\right)^{-1}}\\\
&\qquad=[E^3GIRN^2]\sum_{j=0}^{\infty}\left(\sum_{a\in\mathcal{V}}\frac{a}{1+a}\right)^j\\\
&\qquad=[E^3GIRN^2]\sum_{j=5}^{8}\left(\sum_{a\in\mathcal{V}}\frac{a}{1+a}\right)^j\\\
&\qquad=[E^3GIRN^2]\sum_{j=5}^{8}\left(G+E\left(1-E+E^2\right)\right.\\
&\qquad\qquad\qquad\qquad\qquad\qquad\left.+I+R+N(1-N)\right)^j\\\
&\qquad=-120+1080-3360+3360\\
&\,\,\color{blue}{\qquad=960}
\end{align*}

CALCULATION OF EXPANSION
