How many different words can be formed from the letters of the word DAUGHTER when the letters T,A,D are never together.

Question

How many different words can be formed from the letters of the word DAUGHTER when the letters T,A,D are never together.

Answer 1

I got another answer,it's pretty easy to know that the total of words you can form are $8!$, without any condition. Then, suppose the letters $TAD$ are together and they're one with the $5$ left. Then you have $6!$ ways to write words with $TAD$ together. This is because you can arrange $TAD$ on any of the six spaces, and then the other letters will have $5!$ ways to arrange, then $6*5!=6!$ $$(TAD)UGHER$$$$UG(TAD)HRE$$$$...$$

Remember that you can arrange $TAD$ as $TDA, ADT, ATD, DTA$ and $DAT$, so you have $6!*6$ words with the letters $T,A,D$ together. So you just need to take them away: $$8!-6!*6=40320-4320=36000$$ Hence the answer is $36000$

Answer 2

It is obvious that

-The words can be formed with that letters without any assumption are $8!$

-The letters $T,A,D$ together, can be written with $6$ way $(TAD,TDA,ATD,...)$

We can see that each one of the previous six possibilities can appear in $6$ position, such that the other $5$ letters can be arranged in $5!$ way , so all of them appear $6\times 6 \times 5!$ time, and the answer is ($8!-6\times6!$)