Expanding (2n)!

1. Mar 18, 2013

sandy.bridge

1. The problem statement, all variables and given/known data
Hey all. Not super familiar with using factorials, however, they do pop up occasionally. I understand that n! = 1*2*3*...*n. How do we treat factorial when we are multiplying n by an integer before taking the factorial? I know the answer for expanding (2n)!, however, I do not see why. Thanks in advance.

2. Mar 18, 2013

SteamKing

Staff Emeritus
What don't you see? Think of this:

4! = 4*3*2*1

(2*4)! = 8! = 8*7*6*5*4*3*2*1

3. Mar 18, 2013

sandy.bridge

so (2n)!=(2*1)*(4*3*2*1)*(6*5*4*3*2*1)*...*(2n*(2n-1)*...*1). This can be taken a step further, though correct?

4. Mar 18, 2013

phinds

I don't think you get it. Look at post #2 again.

5. Mar 18, 2013

sandy.bridge

(2n)!=(2n*(2n-1)*(2n-2)*...*1) ?

6. Mar 18, 2013

SammyS

Staff Emeritus
That's more like it!

7. Mar 18, 2013

sandy.bridge

Okay. How exactly does that end up being: 1*2*3*...*n*(n+1)*(n+2)*(n+3)*...*(2n) ?

8. Mar 18, 2013

Ray Vickson

How did (2*4)! = 8! end up being 1*2*3* ... *8? What are you not seeing?

9. Mar 18, 2013

SteamKing

Staff Emeritus
Look at counting to 2n this way:

1,2,3,4,...,n-2,n-1,n - the sequence of all integers from 1 to n
n+1,n+2,n+3,..., n+n-2,n+n-1,n+n - the sequence of integers from n+1 to 2n

10. Mar 18, 2013

skiller

You've written it down correctly twice - one of them is in reverse order of the other. How can you not know that they are the same?

1*2 = 2*1 etc

11. Mar 18, 2013

sandy.bridge

For some reason I was having a hard time seeing from n+1 to 2n. I completely see it now. Thanks!