
Strange factorials
$\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.} $

Re: Strange factorials
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

Re: Strange factorials
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.

Re: Strange factorials
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

Re: Strange factorials
$\displaystyle \\ Hi Pat, \\ 720 \cdot 719 \cdot 718 \cdot ... \cdot 7 = 719! \\ \text{that's a product of only 714 consecutive integers to equal the 719! , we saved 4 multiplications} : lol :$

Re: Strange factorials
Yes ! all this bymber theory stuff has interesting interconnections if only the human mind was capable of understanding it!