maths permutation help An experiment consists of randomly rearranging the 9 letters of the word
TARANTULA,
where all possible orders of the 9 letters are equally likely. Find the probability of each of the following events:
a) 1st 3 letters include no 'A'
In the mark scheme the answer is $\frac{6 \cdot 5 \cdot 4 \cdot 6!}{9!}$ but I thought the denominator should be $\frac{9!}{2! \cdot 3!}$ since there's 3A's and 2T's?
 A: The answer is $$\frac{\frac{6\cdot 5\cdot 4\cdot 6!}{2!3!}}{\frac{9!}{2!3!}}=\frac{6\cdot 5\cdot 4\cdot 6!}{9!}$$There is no $\frac{9!}{2!3!}$ in what you've written simply because the expression is simplified.
The numerator is $\frac{6\cdot 5\cdot 4\cdot 6!}{2!3!}$ because there are $6,5,4$ choices for the first, second, third letters respectively, and then $6,5,4,3,2,1$ choices for the fourth, fifth,..., ninth letters respectively.  
The $T$ repeats twice and the $A$ repeats thrice, so you divide it by $2!3!$.
A: Probability or $P(E)=\frac{Favorable\ Outcomes}{Total \ Outcomes}$ 
Here the total outcomes or Sample Space or $n(S)=\frac{9!}{2!3!}$
Because 9! counts the total permutations and 3! and 2! deletes the redundant permutations of 3 A's and 2 T's.
Now the favorable outcomes would be=$\frac{6.5.4.6.5.4.3.2.1}{2!3!}=\frac{6.5.4.6!}{2!3!}$
 Because $_ _ _ _ _ _ _ _ _$
 To fill the 1st space there are $9-3=6$ characters. Then till the third one there would be $6.5.4$ choices and then for the rest, it would simply be $6!$.  
So $P(E)=\frac{\frac{6.5.4.6!}{2!3!}}{\frac{9!}{2!3!}} $
