binomial coefficient

Physicsissuef

1. The problem statement, all variables and given/known data

How is possible this equality:

[tex] {n \choose k} = \frac{n \cdot (n-1) \cdots (n-k+1)}{k(k-1)...1} = \frac{n!}{k!(n-k)!}[/tex]

? I mean where the second part [tex]\frac{n!}{k!(n-k)!}[/tex] comes from?

2. Relevant equations



3. The attempt at a solution

dirk_mec1

[tex] 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)![/tex]

[tex] k! = k \cdot (k-1) \cdot (k-2) \cdot \cdot \cdot 2 \cdot 1 [/tex]


So we get:

[tex] \frac{n!}{(n-k)!} = n \cdot (n-1) \cdot \cdot \cdot n-k+1 [/tex]

GregA

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

Physicsissuef

Ok, I understand about:

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

dirk_mec1

Note that:

[tex] {n \choose k} \neq \frac{n!}{k!} [/tex]

The definition is the one in your first post.

HallsofIvy

Written like what?

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