- #1
DrummingAtom
- 659
- 2
I saw this in my book and I'm having a difficult time figuring out how this formula was manipulated algebraically to equal the other side. I know that it works because I've used it several times but I can't show that it is true.
[tex] \frac{n(n-1)...(n-r+1)}{r!} = \frac {n!}{(n-r)!r!} [/tex]
One of the things I tried was cross multiplied the r! and canceled it then multilpied the (n-r)!to the other side. Which I was left with:
[tex] [n(n-1)...(n-r+1)]*[(n-r)!]= n! [/tex]
From here I tried expanding the factorials then cancelling stuff but I'm still not seeing the pattern. Thanks for any help.
[tex] \frac{n(n-1)...(n-r+1)}{r!} = \frac {n!}{(n-r)!r!} [/tex]
One of the things I tried was cross multiplied the r! and canceled it then multilpied the (n-r)!to the other side. Which I was left with:
[tex] [n(n-1)...(n-r+1)]*[(n-r)!]= n! [/tex]
From here I tried expanding the factorials then cancelling stuff but I'm still not seeing the pattern. Thanks for any help.