Number of permutations of the word "PERMUTATION" such that no two vowels occur together and no two Ts occur together In how many ways we can arrange the letters of the word "PERMUTATION" such that no two vowels occur together and no two Ts occur together.
I first arranged consonants including one T as below:
$*P*R*M*T*N*$
Now in $6$ star places I will arrange the vowels $A,E,I,O,U$ which can be done in $\binom{6}{5} \times 5!=6!$ ways. Also $P,R,M,N,T$ can themselves arrange in $5!$ ways. Hence, total number of ten letter words now is $5! \times 6!$.
But one $T$ should be placed in eleven places of the ten letter word such that it should not be adjacent to $T$ which is already there. Hence, the remaining $T$ has $9$ ways to place.
Hence, total ways is $6! \times 5! \times 9$.
But my answer is not matching with book answer. Please correct me.
 A: Given that we have two different answers posted thus far (820,800 and 796,800), perhaps I can be forgiven for applying heavy machinery.
We start by replacing all the vowels with Vs and ask how many permutations there are of PVRMVTVTVVN in which no two adjacent letters are equal.  According to [1], the answer is
$$N = \int_0^{\infty} e^{-t}\; \ell_1(t)^4 \;\ell_2(t) \; \ell_5(t) \; dt$$
where $$\begin{align}
\ell_1(t) &= t\\
\ell_2(t) &= \frac{1}{2} t^2 - t\\
\ell_5(t) &= \frac{1}{120}t^5 -\frac{1}{6}t^4 +t^3 -2t^2 + t\\
\end{align}$$
Mathematica evaluates the integral as $N = 6,840$.
To answer the original problem, we multiply by $5!$ to account for all the ways of replacing the five Vs with the five distinct vowels:
$$5! \; N = 820,800$$
[1] Theorem 2.1 in "Counting words with Laguerre series" by Jair Taylor, The Electronic Journal of Combinatorics 21(2), 2014.
http://www.combinatorics.org/ojs/index.php/eljc/article/view/v21i2p1
A: $\require{cancel}$
$\bcancel{* P * R * M * T * T * N *}$
$\bcancel{- First permute the consonants in 6!/2!= 360  ways}$
$\bcancel{In \frac{5}{\binom62} = \frac13 cases, the T's will be together}$
$\bcancel{- Choose and place one vowel between the T's in 5 ways,  
and the balance four in 6\cdot 5\cdot 4\cdot3 = 1600 ways}$
$\bcancel{- For the balance \frac23 place the vowels in 7\cdot6\cdot5\cdot4\cdot3 =2520 ways}$
$\bcancel{- Putting the pieces together, 360[\frac13\cdot 1600 + \frac23\cdot 2520] = 796800}$

Revisiting the question by chance after $7$ years, am aghast to see that my posted answer is  incorrect, though the glitch is mysterious
But you can't argue with Laguerre polynomials, so I decided to counter Laguerre polynomials by Smirnov words, which are words in which no identical letters come together.
Here, I shall lump all the vowels as $V$, and using the formula for Smirnov words, I extract the required coefficient
$[P^1R^1M^1N^1T^2V^5]$ in
$\left(1-\frac{p}{1+p}-\frac{r}{1+r}-\frac{m}{1+m}-\frac{n}{1+n}-\frac{t}{1+t} -\frac{v}{1+v}\right)^{-1}= 6840$
But actually the $V's$ are $5$ distinct vowels, so multiplying by $5!$ we get the correct answer of $\color{blue}{820,800}$
A: Calculate all options without $TT$ restriction and subtract 
the options with two T's together.
$$ 6!/2! * 7 * 6 * 5 * 4 * 3  - 5! * 6*5*4*3*2 =
820800 $$
Slightly different approach:
Add 2 spaces $[]$ as vowels. We must now iterate vowels and consonants, yielding the same sequences as before after dropping the spaces to have a single word.
Since $T$ and $[]$ are duplicate, we have $7!/2 * 6!/2$ solutions.
However, we have to subtract $T[]T$ patterns to prevent two $T$'s together. $T[]T$ patterns can start at 5 places, and we can randomly permute 4 consonants and 6 vowels to fill the rest of the iterating sequence, so subtract $5*4!*6!$.
This solution also results in $820800$, so I have to stick to my solution.
A: The word PERMUTATION has eleven letters, of which six are consonants and five are vowels.  We will first arrange the consonants, then place the vowels.  We will consider two cases, depending on whether the Ts are separated or together in the arrangement of the consonants.
Case 1:  We arrange the six consonants so that the Ts are separated.
There are $4!$ of the consonants P, R, M, N.  This creates five spaces in which to place the Ts.
$$\square C_1 \square C_2 \square C_3 \square C_4 \square$$
To ensure the Ts are separated, we must choose two of these five spaces in which to place the $T$s, which can be done in $\binom{5}{2}$ ways.  We now have seven spaces in which to place the vowels.
$$\square C_1 \square C_2 \square C_3 \square C_4 \square C_5 \square C_6 \square$$
To ensure the vowels are separated, we must choose five of these seven spaces in which to place a vowel, which can be done in $\binom{7}{5}$ ways.  The five vowels can be arranged in the selected spaces in $5!$ ways.  Hence, there are
$$4!\binom{5}{2}\binom{7}{5}5!$$
such arrangements.
Case 2:  We arrange the six consonants so that the Ts are together.
We have five distinct objects to arrange:  P, R, M, N, TT.  They can be arranged in $5!$ ways.  Now that we have arranged the six consonants, we again have seven spaces in which to place the five vowels.  To ensure the Ts are separated, we must place one of the vowels between the two Ts.  To ensure the vowels are separated, we must also place a vowel in four of the remaining six spaces, which can be done in $\binom{6}{4}$ ways.  We can arrange the five vowels in the five selected spaces in $5!$ ways.  Hence, there are
$$5!\binom{6}{4}5!$$
such arrangements.
Total:  Since these cases are mutually exclusive and exhaustive, the number of admissible arrangements is
$$4!\binom{5}{2}\binom{7}{5}5! + 5!\binom{6}{4}5!$$
