In how many ways we can arrange the letters of the word "PERMUTATION" such that no two vowels occur together and no two T's 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
Best Answer
* P * R * M * T * T * N *
First permute the consonants in $6!/2!= 360 $ ways
In $\frac{5}{\binom62} = \frac13$ cases, the T's will be together
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
For the balance $\frac23$ place the vowels in $7\cdot6\cdot5\cdot4\cdot3 =2520$ ways
Putting the pieces together, $360[\frac13\cdot 1600 + \frac23\cdot 2520] = 796800$