Binomial coefficient

Homework Statement

How is possible this equality:

$${n \choose k} = \frac{n \cdot (n-1) \cdots (n-k+1)}{k(k-1)...1} = \frac{n!}{k!(n-k)!}$$

? I mean where the second part $$\frac{n!}{k!(n-k)!}$$ comes from?

The Attempt at a Solution

$$n! = n \cdot (n-1) \cdot (n-2) \cdot \cdot \cdot (n-k-1) \cdot (n-k-2) \cdot \cdot \cdot 3 \cdot 2 \cdot 1 = n \cdot (n-1) \cdot (n-2) \cdot \cdot \cdot (n-(k-1)) \cdot (n-k)!$$

$$k! = k \cdot (k-1) \cdot (k-2) \cdot \cdot \cdot 2 \cdot 1$$

So we get:

$$\frac{n!}{(n-k)!} = n \cdot (n-1) \cdot \cdot \cdot n-k+1$$

multiply the first bit by (n-k)!/(n-k)!

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$$\frac{n!}{(n-k)!} = n \cdot (n-1) \cdot \cdot \cdot n-k+1$$
But still I can't understand why n! is written like that..

$$\frac{n!}{(n-k)!} = n \cdot (n-1) \cdot \cdot \cdot n-k+1$$
But still I can't understand why n! is written like that..

Note that:

$${n \choose k} \neq \frac{n!}{k!}$$

The definition is the one in your first post.

HallsofIvy
Homework Helper

$$\frac{n!}{(n-k)!} = n \cdot (n-1) \cdot \cdot \cdot n-k+1$$
But still I can't understand why n! is written like that..
Written like what?

You started by saying that you understand that
$${n \choose k} = \frac{n \cdot (n-1) \cdots (n-k+1)}{k(k-1)...1}$$
but did not understand why
$$\frac{n \cdot (n-1) \cdots (n-k+1)}{k(k-1)...1}= \frac{n!}{k!(n-k)!}$$
That is what was just explained. It is written "that way" in order to give a simple closed form expression to
$$\frac{n \cdot (n-1) \cdots (n-k+1)}{k(k-1)...1}$$
"without the dots".