## binomial coefficient

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

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?

2. Relevant equations

3. The attempt at a solution

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

## binomial coefficient

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

 Quote by 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..
Note that:

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

The definition is the one in your first post.

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 Quote by 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..
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".