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:

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

 

[dirk_mec1]

Ok, I understand about:

<br /> \frac{n!}{(n-k)!} = n \cdot (n-1) \cdot \cdot \cdot n-k+1 <br />
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:

<br /> \frac{n!}{(n-k)!} = n \cdot (n-1) \cdot \cdot \cdot n-k+1 <br />
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".