Understanding the Equation (2n-1)! = (2n-1)(2n)(2n-1)!

[fiziksfun]

Can someone please explain to me why (2n-1)! = (2n-1)(2n)(2n-1)! ?? I'm very confused.

 

[marcusl]

It doesn't. Did this come from a book?

 

[gabbagabbahey]

Do you mean (2n)!= (2n)(2n-1)! ? If so, it comes from the definition of factorial: (2n)!=(2n)(2n-1)(2n-2)...(1)
and (2n-1)!=(2n-1)(2n-2)...(1) so (2n)(2n-1)!=(2n)(2n-1)(2n-2)...(1)=(2n)!

 

[HallsofIvy]

Can someone please explain to me why (2n-1)! = (2n-1)(2n)(2n-1)! ?? I'm very confused.

Yes, you are! Dividing both sides of your formula by (2n-1)! you get 1= (2n-1)(2n) which is NOT true!

Perhaps you mean (2n+1)!= (2n+1)(2n)(2n-1)!. That's true because, by definition, (2n+1)!= (2n+1)(2n)(2n-1)(2n-2)(2n-3)(2n-4)...(3)(2)(1). And (2n-1)!= (2n-1)(2n-3)(2n-4)...(3)(2)(1), the "tail end" of that first product. so (2n+1)!= (2n+1)(2n)(2n-1)!