Operating room schedule probability (Statistics homework problem) I am trying to figure out the following homework problem from the textbook Applied Statistics and Probability for Engineers (6th Edition):
Question: Suppose that an operating room needs to schedule 2 knee, 4 hip, and 5 shoulder surgeries. Assume that all schedules are equally likely.
Find the probability that the schedule begins with a hip surgery, given that all of the shoulder surgeries are last.

Attempted solution:
If A is the event that a schedule begins with a hip surgery, then we have 3 other hip surgeries remaining.
Let B is the event that shoulder surgeries are last.
$$
\begin{align}
P(A|B) & =\frac{P(A \cap B)}{P(B)}\\
& =\frac{\frac{5!}{1!4!}}{\frac{6!}{2!4!}}\\
& =\frac{5}{15}\\
& =\frac{1}{3}
\end{align}
$$
Does my attempted solution make sense? The number of schedule permutations in (A $\cap$ B) is $\frac{5!}{1!4!}$ because there are 5 operations that can be freely scheduled and we have 1 knee and 4 hip operations left to schedule. The number of schedule permutations in (B) is $\frac{6!}{2!4!}$ because there are 6 operations to schedule besides the shoulder operations and we have 2 knee and 4 hip operations left to schedule.
$\therefore P(A|B) = \frac{1}{3}$.
 A: 
The number of schedule permutations in $(A ∩ B)$ is $\frac{5!}{1!4!}$ because there are $5$ operations that can be freely scheduled and we have $1$ knee and $4$ hip operations left to schedule. 

First of all, permutation implies ordered operations. Secondly, the number of scheduled permutations in $(A\cap B)$ is not equal to $P(A\cap B)$, because the latter is probability. Thirdly, the number of scheduled permutations in $(A\cap B)$ is not $\frac{5!}{1!4!}$ (it is ${4\choose 1}\cdot 5!\cdot 5!$). 
Your thought: $5$ operations to be freely scheduled: $K,H,H,H,H$, so $\frac{5!}{1!4!}$. However, you must consider $K,H_1,H_2,H_3,H_4$ and $KH_1H_2H_3H_4$ is different from $KH_2H_1H_3H_4$. Moreover, you must look at the general picture. 
Denote: $K_1,K_2,H_1,H_2,H_3,H_4,S_1,S_2,S_3,S_4,S_5$ the knee, hip and shoulder operations, respectively. 
First, allocate $5$ shoulder operations to the last $5$ positions, which can be done in $\color{green}{5!}$ ways.
Second, select $1$ hip operation to the first position, which can be done in $\color{red}{{4\choose 1}}=4$ ways. (Note that here you are also confusing with knee operation)
Third, once $5$ shoulder operations are set to the last and $1$ hip operation is set to the first positions, there remained overall $5$ operations ($2$ knee and $3$ hip) for the middle $5$ positions, which can be arranged in $\color{blue}{5!}$ ways.
Hence, the number of scheduled permutations in $A\cap B$ is:
$$\color{red}{{4\choose 1}}\cdot \color{green}{5!}\cdot \color{blue}{5!}$$
Similarly, the number of outcomes in $B$ is:
$$5!\cdot 6!$$
because, there are $5!$ ways to arrange $5$ shoulder operations and then $6!$ ways to arrange the rest $6$ ($2$ knee and $4$ hip) operations.
Hence, the required probability is:
$$\frac{{4\choose 1}\cdot 5!\cdot 5!}{6!\cdot 5!}=\frac23.$$
Alternatively, since it is stated "given that all of the shoulder surgeries are last", the shoulder operations can be ignored and only $2$ knee and $4$ hip operations can be considered for having a hip operation first, which is:
$$\frac{{4\choose 1}\cdot 5!}{6!}=\frac23.$$ 
A: You've written your conditional probability as a ratio of other probabilities, but then you somehow switch to counting outcomes, rather than probabilities without justification.  For instance, your numerator is $5!/(1! 4!)$ but this is not a probability; neither is the denominator $6!/(2! 4!)$.  So you are missing some crucial details of your explanation.
Moreover, you are counting things without taking into consideration the other operations.  In other words, you aren't actually counting elementary outcomes.
To count correctly, we must consider all elementary outcomes for which the desired ordering occurs, versus all elementary outcomes that can occur.
The former comprises outcomes in which one of the four hip surgeries is first, and all five shoulder surgeries are last.  The first surgery can be chosen in $4$ ways; the last five surgeries can be chosen in $5!$ ways.  In between, there are $3 + 2 = 5$ other surgeries that may be scheduled without regard to order; thus the total number of desired outcomes is $$4 \cdot 5! \cdot 5! = 56700.$$  Each one is equally likely to occur because we are told each possible ordering (of which there are $11!$) is equally likely.
The latter comprises outcomes in which all five shoulder surgeries are last.  The other six surgeries can be chosen in any order.  This can be done in $$6! \cdot 5! = 86400$$ ways.  Therefore, the desired probability is $$\frac{4 \cdot 5! \cdot 5!}{6! \cdot 5!} = \frac{2}{3}.$$
Note that your answer of $1/3$ is the probability that the first surgery is a knee surgery given that the last five are all shoulder surgeries.  Your question asks for the probability that the first surgery is a hip surgery given that the last five are all shoulder surgeries.  Also note that the counting method described in the comments is much simpler and equally valid (although it seems the roles of knee and hip got reversed); however, I have chosen to explain a more detailed enumeration method to illustrate how you would actually go about enumerating the elementary outcomes, each of which comprises a full ordering or permutation of the eleven surgeries scheduled.
