I'm having difficulty with the following problem:
You choose a letter at random from the word Mississippi eleven times
without replacement. What is the probability that you can form the
word Mississippi with the eleven chosen letters? Hint: it may be
helpful to number the eleven letters as $1, 2, . . . , 11$.
This is how I approached it: the number of possible outcomes is $11!$, since we're taking one word at a time without replacement, so there are $11$ choices for the first letter, $10$ choices for the second letter and so on. Now, as to the number of 'success' outcomes, I noticed that there are $4$ s's to choose from, $4$ i's, $1$ M and $2$ p's. So, in order to form the word Mississippi, we have for the first letter $1$ option, $4$ for the second and third letters, $3$ for the fourth (since we've already used one "s") and so on, which amounts to a total of $4^2*3^2*2^3=1152$ different ways of doing so.
However, my answer does not match the one provided in my book (Henk Tijn's Understanding Probability 3rd edition). What am I doing wrong? Thanks very much in advance.
Best Answer
There are $11!$ permutations of 11 letters but the order of the 4 s's, 4 i's and 2 p's doesn't matter. This means there are $2! 4! 4!$ indistinguisable permutations for any permutation of the 11 letters. Therefore there are \begin{equation} \frac{11!}{2! 4! 4!} = 34650 \end{equation} ways of arranging the letters of Mississippi, making the probability $1/34650$ that a random permuatation spells Mississippi.