What is the origin of the binomial coefficient formula equality?

[Physicsissuef]

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?

Homework Equations

The Attempt at a Solution

 

[dirk_mec1]

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

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


So we get:

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

 

[GregA]

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

 

[Physicsissuef]

Ok, I understand about:

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

 

[dirk_mec1]

Ok, I understand about:

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

Ok, I understand about:

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