Strange factorials

Aug 2011
105
42
42nd parallel, North America
\(\displaystyle \\ 10 \cdot 9 \cdot 8 \cdot 7 = 7! \\ \\ 6 \cdot 5 \cdot 4 = 5! \\ \\ \text{Does anybody know or can generate other cases like this? Where you have a product of consecutive integers, not including 2,} \) \(\displaystyle \\ \text{and this product is equal to the factorial of one of the integers used in the product.} \)
 
Nov 2012
4
1
UK
that's cool never thought of before. there are an infinite number of similar examples and they are easy to construct. works like this consider an integer n. now call n! = k. now find k!. we know (k- 1)! = k!/k = k!/n!

that's it really or more generally (n!-1)! = (n!)!/n!

try it out it works

nice one

pat
 
  • Like
Reactions: 1 person
Aug 2011
105
42
42nd parallel, North America
I like your solution, thank you for the reply, let me put it in my own words. 'We can always generate this kind of relationship if we take the factorial of a factorial.' Interesting that \(\displaystyle \frac{10!}{6!} = 7! \) falls outside this generating pattern.
 
Nov 2012
4
1
UK
An example might be good...
1.2.3.4 = 24.
so if i include 1.2.3.4 in one series and exclude 24. Then in the other i include 24 but exclude 1.2.3.4

ie 5.6.7.8.9....23. 24 = 1.2.3.4.5.6....22.23

That makes sense.
because
In your is slightly different 6.5.4 = 5.4.3.2.1 because 4.6 = 1.2.3.4

we try 1.2.3.4.5 = 120 = 5.24

so 1.2.3.4.5.6.7.....23 = 5.6.7.8.9....23.24

i suppose there must be loads of ways to do this.
All very interesting though, I never really thought about that before.

Pat
 
Aug 2011
105
42
42nd parallel, North America
\(\displaystyle \\ Hi Pat, \\ 720 \cdot 719 \cdot 718 \cdot ... \cdot 7 = 719! \\ \text{that's a product of only 714 consecutive integers to equal the 719! , we saved 4 multiplications} : lol :\)
 
Last edited:
Nov 2012
4
1
UK
Yes ! all this bymber theory stuff has interesting interconnections if only the human mind was capable of understanding it!