<s>[INST] 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?
 [/INST] 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 $2!3!$.
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