1
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ Please see this tutorial on how to typeset mathematics on this site. $\endgroup$ – N. F. Taussig Apr 13 '15 at 18:50
4
$\begingroup$

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 $2!3!$.

$\endgroup$
0
$\begingroup$

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!}} $

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.