Probability of number of drawing cards in a scenario being equal to that in another scenario I came across the following question in a book:-
$Q.$ Cards are drawn one by one at random from a well shuffled pack of $52$ cards.
$(a)$Find the probability that exactly $n$ cards are drawn before the first king appears.
$(b)$If $N$ is the number of cards required to be drawn until $2$ aces are obtained for the first time then show that probability of $N$ being equal to $n$ is $$\frac {(n-1)(52-n)(51-n)}{50*49*17*13} \space \space where \space 2\le n\le 50 .$$
The first part is easy. It is simply that no king has been chosen in the last $n$ cards. At the $(n+1)th$ card, we have $4$ kings to choose from. Hence $$P(n)=\frac {\binom {48}{n}\binom{4}{1}}{\binom {52}{n}}=\frac {(52-n)(51-n)(50-n)(49-n)}{13*51*50*49}$$ 
For the second part we now that the last card drawn is the second ace (which can be chosen in $3$ ways as one ace has already been chosen) and none of the cards drawn before is a king (as $N=n$). Hence probability of drawing the aces is $$P(N)=\frac {3\binom{4}{1}\binom{44}{n-1}}{\binom{52}{n}}$$ But how do I find the probability of $n$ being equal to $N$?
 A: (a) Andre's comment is right about needing to include the king in your calculations.
There are:
\begin{eqnarray*}
&&\text{$\binom{48}{n}$ ways to choose the first $n$ cards} \\
&&\text{$\binom{4}{1}$ ways to choose the first king} \\
&&\text{$n!$ ways to arrange those first $n+1$ cards} \\
&&\text{$(51-n)!$ ways to arrange the last $51-n$ cards} \\
&&\text{$52!$ ways to arrange all $52$ cards.} \\
\end{eqnarray*}
Therefore,
\begin{eqnarray*}
P(\text{$n$ cards before first king}) &=& \dfrac{\binom{48}{n}\binom{4}{1}n!(51-n)!}{52!} \\
&& \\
&=&\dfrac{(51-n)(50-n)(49-n)}{13\cdot51\cdot50\cdot49}\qquad\qquad\text{after simplifying.} \\
\end{eqnarray*}
$$\\$$
(b) Here, there are:
\begin{eqnarray*}
&&\text{$\binom{48}{n-2}$ ways to choose the first $n-2$ non-ace cards} \\
&&\text{$\binom{4}{2}$ ways to choose the first two aces} \\
&&\text{$2!(n-2)!$ ways to arrange those first $n$ cards} \\
&&\text{$(52-n)!$ ways to arrange the last $52-n$ cards} \\
&&\text{$52!$ ways to arrange all $52$ cards.} \\
\end{eqnarray*}
Therefore,
\begin{eqnarray*}
P(\text{$n^{th}$ card is the second ace}) &=& \dfrac{\binom{48}{n-2}\binom{4}{2}2!(n-2)!(52-n)!}{52!} \\
&& \\
&=&\dfrac{(n-1)(52-n)(51-n)}{50\cdot49\cdot17\cdot13}\qquad\qquad\text{after simplifying.} \\
\end{eqnarray*}
