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 $$\frac{14!}{4! 8}-\frac{10!}{4}-\frac{11!}{6}+\frac{8!}{4}+7!+\frac{8!}{2}=446528880.$$