Calculate number of options with no repetitions Given the word "REPETITIVENESS" calculate the number of words that does not include the series "SITE" OR "TERSE"
For example:
REPETIVSITEENS is not valid
REPETITIVENESS is valid
I tried to count each letter:
R-1
E-4
P-1
T-2
I-2
V-1
N-1
S-2
and so the total permutations are: $\frac{14!}{4! \cdot 2! \cdot 2! \cdot 2!} = 454053600 $ and then use the include exclude theorem {permutations of words including SITE} and {perm. of words including TERSE}
but it seems incorrect... I don't know how to solve it I would like your help. Thank you!!

We already know the number of total arrangements,
$$\frac{14!}{4! \cdot 2! \cdot 2! \cdot 2!}=\frac{14!}{4! 8}.$$
The number of words including TERSE is
$$10\cdot\frac{9!}{2!2!}=\frac{10!}{4}.$$
The number of words including SITE (at least one time) is
$$11\cdot\frac{10!}{3!}-\binom{6+2}{2}\cdot\frac{6!}{2!}=
\frac{11!}{6}-\frac{8!}{4}.
$$
The number of words including SITE and TERSE as distinct words is
$$2\cdot \binom{5+2}{2}\cdot 5!=7!.$$
The number of words including SITERSE is
$$8\cdot \frac{7!}{2!}=\frac{8!}{2}.$$
Hence, by the inclusion-exclusion principle, we find
$$\frac{14!}{4! 8}-\frac{10!}{4}-\frac{11!}{6}+\frac{8!}{4}+7!+\frac{8!}{2}=446528880.$$
