- #1

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## Homework Statement

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?

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- Thread starter Physicsissuef
- Start date

- #1

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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

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[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]

- #3

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

Last edited:

- #4

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[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..

- #5

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[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..

Note that:

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

The

- #6

HallsofIvy

Science Advisor

Homework Helper

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Written like

[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..

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".

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