# Notation for Factorials!

• I
Staff Emeritus
I just have a quick question on how to write the notation for a factorial. I have a series with a factorial of 5*10*15*...*(5n) in it. Is this written as 5n!, as (5n)!, or something else? I'm pretty sure it's 5n!, as I've written 5n! out as 5(1*2*3*4*...*n), which when you distribute the 5 appears to come out as (5*10*15*...*5n), but I just wasn't sure if I'd broken some math rule somewhere.

Also, if it is 5n! and not (5n)!, can (5n)! be easily expressed in a form similar to (1*2*3*...*n)?

Thank you!

which when you distribute the 5 appears to come out as (5*10*15*...*5n)

Multiplication doesn't distribute over multiplication.

I think you're looking for a quintuple factorial, i.e. ##(5n)!!!!!##. For instance, the double factorial is defined as $$(n)!!=n \cdot (n-2) \cdot (n-4) ... 1$$

Drakkith
Staff Emeritus
Multiplication doesn't distribute over multiplication.

Hah! Of course it doesn't! Silly me!
@phinds You're rubbing off on me, old man!

I think you're looking for a quintuple factorial, i.e. (5n)!!!!!(5n)!!!!!(5n)!!!!!. For instance, the double factorial is defined as
(n)!!=n⋅(n−2)⋅(n−4)...1(n)!!=n⋅(n−2)⋅(n−4)...1​
(n)!!=n \cdot (n-2) \cdot (n-4) ... 1

Thanks! I'll look into it!

Mark44
Mentor
I just have a quick question on how to write the notation for a factorial. I have a series with a factorial of 5*10*15*...*(5n) in it
This would be ##5^n(1 * 2 * 3 * ... * n)## or ##5^n * n!##. Each of the n factors in the original expression has a factor of 5, which gives the ##5^n## part, and the remaining part is 1 * 2 * 3 * ... * n, or n!.
Drakkith said:
. Is this written as 5n!, as (5n)!, or something else? I'm pretty sure it's 5n!, as I've written 5n! out as 5(1*2*3*4*...*n), which when you distribute the 5 appears to come out as (5*10*15*...*5n), but I just wasn't sure if I'd broken some math rule somewhere.

Also, if it is 5n! and not (5n)!, can (5n)! be easily expressed in a form similar to (1*2*3*...*n)?

Thank you!

Staff Emeritus
This would be 5n(1∗2∗3∗...∗n)5n(1∗2∗3∗...∗n)5^n(1 * 2 * 3 * ... * n) or 5n∗n!5n∗n!5^n * n!. Each of the n factors in the original expression has a factor of 5, which gives the 5n5n5^n part, and the remaining part is 1 * 2 * 3 * ... * n, or n!.

Gah! Somehow I missed the notification that you replied last night, Mark. I was just about to post the correct notation, which is 5nn!. Just got help from a tutor here on campus who figured it out. It all makes perfect sense now!

ProfuselyQuarky
Gold Member
All these exclamation marks . . . everyone is so excited!!!!