smc

We consider the vowels and consonants separately. There are $2$ vowels ($O$ and $E$), giving $2!=2$ choices for the first two letters; similarly, there are $5$ consonants ($C$, $N$, $S$, and two $T$s), which would give $5!=120$ possible choices for letters $3$ to $7$, except that since the two $T$s are indistinguishable, this actually counts each order exactly twice. Therefore the number of possible orderings of the consonants is $\frac{120}{2}=60$, giving a total of $2\cdot 60=\boxed{120}$ possible rearrangements.