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?

  • $\begingroup$ Please see this tutorial on how to typeset mathematics on this site. $\endgroup$ – N. F. Taussig Apr 13 '15 at 18:50

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 $2!3!$.


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=$\frac{}{2!3!}=\frac{!}{2!3!}$
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{!}{2!3!}}{\frac{9!}{2!3!}} $


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