Could someone please explain this to me

[Pojebany]

Hi, could someone please explain how this: [MATH]n!/2!(n-2)![/MATH] gets to this: [MATH]n(n-1)/2[/MATH], thanks

 

[lev888]

Hi, could someone please explain how this: [MATH]n!/2!(n-2)![/MATH] gets to this: [MATH]n(n-1)/2[/MATH], thanks

Expand the factorials and you'll see that most of the terms cancel.

 

[Deleted member 4993]

Hi, could someone please explain how this: [MATH]n!/2!(n-2)![/MATH] gets to this: [MATH]n(n-1)/2[/MATH], thanks

1. Write the first four terms of n!, starting from n = ?
   
   
2. Write the first four terms of (n-2)! starting from (n-2) = ?

What do you get when you divide (1) by (2)?

 

[lookagain]

Hi, could someone please explain how this: thanks

Hi, could someone please explain how this: [MATH]n!/2!(n-2)![/MATH] gets to this: [MATH]n(n-1)/2[/MATH], thanks

You have to type the first denominator in grouping symbols, because as written, it does not mean as you intended:

[MATH]n!/[2!(n-2)!][/MATH] gets to this: [MATH]n(n-1)/2[/MATH]

 

[pka]

Hi, could someone please explain how this: [MATH]n!/2!(n-2)![/MATH] gets to this: [MATH]n(n-1)/2[/MATH]

I think the best way to learn about these is to do some: consider $\displaystyle \dfrac{7!}{5!}=\dfrac{7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1}{5\cdot4\cdot3\cdot2\cdot1}=7\cdot 6$ Do you see that?

Now consider $\displaystyle \dfrac{7!}{2!(7-2)!}=\dfrac{7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1}{2!(5\cdot4\cdot3\cdot2\cdot1)}=\dfrac{7\cdot 6}{2!}$

Does it now make sense that $\displaystyle \dfrac{n!}{2!(n-2)!}=\dfrac{n\cdot(n-1)}{2!}~~?$

 

[Pojebany]

Thanks for the explanations, makes sense now.