The probability of forming Mississippi by choosing random letters from Mississippi 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. 
 A: 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.
A: Here is the word $$\begin{matrix}M&I&S&S&I&S&S&I&P&P&I\\1&2&3&4&5&6&7&8&9&10&11\end{matrix}$$with a numbering of the drawings.
1 The probability that you choose letter $M$ the first time is $\frac{1}{11}.$
2 Then, given 1, the probability that you choose letter $I$ the second time is $\frac4{10}$. 
From this point on the phrase: "given the result of the previous drawings" will be omitted.
3 The probability that you choose letter $S$ the third time is $\frac49$. 
4 The probability that you choose letter $S$ the fourth time is $\frac38$.
5 The probability that you choose letter $I$ the fifth time is $\frac 37$.
6 The probability that you choose letter $S$ the sixth times  is $\frac26$.
7 The probability that you choose letter $S$ the seventh times  is $\frac15$.
8 The probability that you choose letter $I$ the eighth time is $\frac24$.
9 The probability that you choose letter $P$ the ninth time is $\frac23$.
10The probability that you choose letter $P$ the tenth time is $\frac12$.
11The probability that you choose letter $I$ the eleventh time is $1$.
So, the probability that you get the word again is
$$\frac{1}{11}\frac4{10}\frac49\frac38\frac 37\frac26\frac15\frac24\frac23\frac12=\frac{2(4!)^2}{11!}$$
