Understanding Factorials and Multiplying by an Integer

[sandy.bridge]

Homework Statement

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.

 

[SteamKing]

What don't you see? Think of this:

4! = 4*3*2*1

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

 

[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?

 

[phinds]

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?

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

 

[sandy.bridge]

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

 

[SammyS]

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

That's more like it!

 

[sandy.bridge]

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

 

[Ray Vickson]

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

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

 

[SteamKing]

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

 

[skiller]

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

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

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

 

[sandy.bridge]

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

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