Finding the combination I have this formula: 
$$\frac{(N-K)(N-K-1)\cdots(N-K-n+1)}{N(N-1)\cdots(N-n+1)}$$
and my book says it can be written like this:
$$
\frac{\binom {N-n}K}{\binom N K}
$$
The problem is i cannot understand how they got to that result. I have been trying to calculate my way back from it to my original formula, but it doesn't give me the right result.
Please be patient with me, I am a beginner. And sorry that I've put it in as an image, i don't know how to write it here.
Update: My formula comes from calculating draws, without replacement:
I have $N$ balls, of which $K$ are white and I want to calculate the probability of not getting a white one in the first $n$ draws.
So $$
\begin{align}
P(1) & = \frac{N-K}{N} \\[6pt]
P(2) & = \frac{(N-K)(N-K-1)}{N(N-1)} \\[6pt]
P(n) & = \frac{(N-K)(N-K-1)\cdots(N-K-n+1)}{N(N-1)\cdots(N-n+1)}
\end{align}
$$
I am following a paper on the subject, and so far I was understanding, but I got stuck at 
$$
\frac{\binom {N-n}K}{\binom N K}
$$
 A: Here is a way to see that the two expressions are equal without going through all the factorial computations.
As the OP notes, the first displayed formula, which can be thought of as written in the form
$${N-K\over N}\cdot{N-K-1\over N-1}\cdot{N-K-2\over N-2}\cdots{N-K-{n-1}\over N-(n-1)}$$
calculates the probability that you do not choose any of the $K$ white balls in the first $n$ draws when drawing without replacement from among $N$ balls.  But another way to compute this probability is to imagine lining up $N$ uncolored balls in the order in which they'll be drawn (not stopping after $n$ draws) and then picking $K$ balls at random to be colored white.  The total number of ways to pick $K$ balls is $N\choose K$; the number of ways to do so while avoiding the first $n$ balls is $N-n\choose K$.  So the probability is the quotient of those, or
$${N-n\choose K}\over{N\choose K}$$
A: It does seem as if your second formula may have a typo since
$\require{cancel}$
$$\xcancel{
\begin{align}
\frac{(N-K)(N-K-1)\cdots(N-K-n+1)}{N(N-1)\cdots(N-n+1)}
&=\frac{\frac{(N-K)!}{(N-K-n)!}}{\frac{N!}{(N-n)!}}\\[4pt]
&=\frac{\frac{(N-K)!}{(N-K-n)!\,n!}}{\frac{N!}{(N-n)!\,n!}}\\[4pt]
&=\frac{\binom{N-K}{n}}{\binom{N}{n}}
\end{align}}
$$

Seeing Barry Cipra's comment, I feel quite foolish. I have used this identity in many answers. Here is a corrected answer.
$$
\begin{align}
\frac{(N-K)(N-K-1)\cdots(N-K-n+1)}{N(N-1)\cdots(N-n+1)}
&=\frac{\frac{\color{#C00000}{(N-K)!}}{(N-K-n)!}}{\frac{N!}{\color{#00A000}{(N-n)!}}}\\[4pt]
&=\frac{\frac{\color{#00A000}{(N-n)!}}{(N-K-n)!}}{\frac{N!}{\color{#C00000}{(N-K)!}}}\\[4pt]
&=\frac{\frac{(N-n)!}{(N-K-n)!\,K!}}{\frac{N!}{(N-K)!\,K!}}\\[4pt]
&=\frac{\binom{N-n}{K}}{\binom{N}{K}}
\end{align}
$$
A: Hint:
$$N(N-1)\cdots(N-n+1) = \dfrac{N!}{(N-n)!} = n!{N \choose n}$$
You may have a typo in your original question, confusing $n$ and $K$
A: $\dbinom{N-n}{K}=\dfrac{(N-n)!}{K![K-(N-n)]!}$ & $\dbinom{N}{K}=\dfrac{N!}{K!(K-N)!}$.. taking ratio of these $\frac {\dbinom{N-n}{K}}{\dbinom{N}{K}}=\frac{\dfrac{(N-n)!}{K![K-(N-n)]!}}{\dfrac{N!}{K!(K-N)!}}$=$\dfrac{(N-n)! K!(K-N)!}{K!N![K-(N-n)]!}$=$\dfrac{(N-n)! (K-N)!}{N![(K-N)+n]!}$.... Hope you can do it from here.
A: just use the combination formula for \begin{pmatrix}N-n\\K\end{pmatrix}and \begin{pmatrix}N\\K\end{pmatrix} separately and do the simplification. Go from R.H.S. to L.H.S. or say answer to question.
That will be easier to understand.
