The discussion revolves around understanding factorials, particularly how to handle the factorial of a product involving an integer, such as (2n)!. The original poster expresses confusion about the relationship between factorials and multiplication by an integer, specifically in the context of expanding (2n)!. Participants are exploring the definitions and properties of factorials in this context.
The discussion is active, with participants providing insights and clarifications. Some have offered guidance on understanding the sequences leading to (2n)!, while others are still grappling with the conceptual connections. There appears to be a productive exchange of ideas, with some participants expressing newfound clarity.
There is a noted difficulty in visualizing the sequence of integers from n+1 to 2n, which has been a point of confusion for some participants. The original poster's familiarity with factorials is limited, which may be influencing the discussion dynamics.
sandy.bridge said: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?
That's more like it!sandy.bridge said:(2n)!=(2n*(2n-1)*(2n-2)*...*1) ?
sandy.bridge said:Okay. How exactly does that end up being: 1*2*3*...*n*(n+1)*(n+2)*(n+3)*...*(2n) ?
sandy.bridge said:(2n)!=(2n*(2n-1)*(2n-2)*...*1) ?
sandy.bridge said:Okay. How exactly does that end up being: 1*2*3*...*n*(n+1)*(n+2)*(n+3)*...*(2n) ?
For some reason I was having a hard time seeing from n+1 to 2n. I completely see it now. Thanks!SteamKing said: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