What is the probability that the $k$-ith inspected piece be the last defective piece 
A set with $n$ pieces has $r$ defective pieces. If the order of inspection is randomly done, what is the probability that the $k$-ith inspected piece ($k\geq r $) be the last defective piece in the set?

What i've done:
Let $P_1,P_2,...,P_{k-1},P_k, P_{k+1},...,P_n$ be the sequence of pieces satisfying the hyphotesis, for $k\geq r$.
In order to $P_k$ be the last defective piece, we must distribute $r-1$ defective pieces into the $k-1$ first positions. The $k-r$ remaining positions into the $k-1$ first positions must be filled with satisfying pieces. Also, the $n-k$ last positions after the $k$-ith must be filled with working pieces.
Therefore, for the $k-1$ first positions, there are ${\binom{r}{r-1}\binom{n-r}{k-r}}$
possible ways to choose those pieces.
I know that dividing by $\binom{n}{k-1}$ would give me the probability to the $k$-ith position be the last piece of $k$ pieces. But for $n$ pieces, my book's answer also divides this result by $1/(n-k+1)$ and i don't know why this would give me the desired probability.
Can someone give any explanation?
 A: I have managed to find my own answer. Here it is:
Let $P_1,P_2,...,P_{k-1},P_k, P_{k+1},...,P_n$ be the sequence of pieces satisfying the hyphotesis.
In order to $P_k$ be the last defective piece, we must distribute $r-1$ defective pieces into the $k-1$ first positions. The $k-r$ remaining positions into the $k-1$ first positions must be filled with satisfying pieces. We also notice that there are $n-k+1$ possible positions to put the last defective piece into the $n-k+1$ remaining pieces. This gives us $n-k$ positions to fill up with working pieces until the last one. 
Therefore, there are $\binom{r}{r-1}$ ways to choose $r-1$ defective pieces from a group containing $r$ of them, $\binom{n-r}{k-r}$ ways to choose $k-r$ working pieces to put into the remaining $k-1$ first positions and $\binom{n}{k-1}$ ways to choose $k-1$ pieces from a set with $n$ pieces. And since there are $n-k+1$ possible ways to put the last defective one, the desired probability is:
$$P=\frac{\binom{r}{r-1}\binom{n-r}{k-r}}{\binom{n}{k-1}(n-k+1)}$$
(i'm sorry for my english)
