Function Posted March 15, 2014 Share Posted March 15, 2014 (edited) Hello everyone I played around with some binomials, variations etc. and I stumbled upon a problem. Let's first of all notice the following: [math]V^a_b=\frac{b!}{(b-a)!}=\binom{b}{a}\cdot a![/math] An equation which I have proven in "Variations formula", a topic of mine: [math]V^{n-m}_n\cdot(n+1)=V^{n-m+1}_{n+1}[/math] [math]\Leftrightarrow \binom{n}{n-m}(n-m)!(n+1)=\binom{n+1}{n-m+1}(n-m+1)![/math] [math]\Leftrightarrow \binom{n}{n-m}(n-m)!(n+1)=\binom{n+1}{n-m+1}(n-m+1)(n-m)![/math] [math]\Leftrightarrow \binom{n}{n-m}(n-m)!(n+1)=(n+1)\binom{n+1}{n-m+1}-m\binom{n+1}{n-m+1}[/math] [math]\Leftrightarrow \left(\binom{n}{n-m}-\binom{n+1}{n-m+1}\right)(n+1)=-m\binom{n+1}{n-m+1}[/math] [math]\Leftrightarrow -\binom{n}{n-m}(n+1)=-m\binom{n+1}{n-m+1}[/math] [math]\Leftrightarrow \frac{-n!}{(n-m)!m!}(n+1)=-m\cdot\frac{(n+1)!}{(n-m+1)!m!}[/math] [math]\Leftrightarrow \frac{-n!(n+1)}{(n-m)!m!}=m\cdot\frac{-n!(n+1)}{(n-m+1)!m!}[/math] [math]\Leftrightarrow \frac{m}{(n-m+1)!}=\frac{1}{(n-m)!}[/math] [math]\Leftrightarrow \frac{m}{(n-m+1)(n-m)!}=\frac{1}{(n-m)!}[/math] [math]\Leftrightarrow \frac{m}{n-m+1}=1[/math] Obviously, this isn't always true. I can't seem to find any mistake, however. Can someone help me on this one? Thanks. Function Edited March 15, 2014 by Function Link to comment Share on other sites More sharing options...

John Posted March 15, 2014 Share Posted March 15, 2014 (edited) [math]\Leftrightarrow \binom{n}{n-m}(n-m)!(n+1)=\binom{n+1}{n-m+1}(n-m+1)(n-m)![/math] [math]\Leftrightarrow \binom{n}{n-m}(n-m)!(n+1)=(n+1)\binom{n+1}{n-m+1}-m\binom{n+1}{n-m+1}[/math] You seem to have forgotten to factor out (n-m)! on the left side here, but I guess that's just a typo since you seem to continue as if you had factored it out. [math]\Leftrightarrow \left(\binom{n}{n-m}-\binom{n+1}{n-m+1}\right)(n+1)=-m\binom{n+1}{n-m+1}[/math] [math]\Leftrightarrow -\binom{n}{n-m}(n+1)=-m\binom{n+1}{n-m+1}[/math] I don't think these two equations are equivalent. Consider n = 2 and m = 1. Then [math] {2\choose1} - {3\choose2} = 2 - 3 = -1[/math] while [math]-{2\choose1} = -2[/math]. Edit: And just as a general piece of advice, you might want to learn how to align equations in LaTeX (if you don't know already). Separating everything out made this OP somewhat hard to follow, especially on the mobile site where the different formula lengths meant the font size varied from line to line. It's not a huge deal, I guess, but for strings of equations this long, the added clarity would be nice. Edited March 15, 2014 by John Link to comment Share on other sites More sharing options...

Function Posted March 15, 2014 Author Share Posted March 15, 2014 I don't think these two equations are equivalent. Consider n = 2 and m = 1. Then [math] {2\choose1} - {3\choose2} = 2 - 3 = -1[/math] while [math]-{2\choose1} = -2[/math]. Ah yes.. I transformed a proven equation wrongly. My bad. I'll look into it tomorrow. Thanks. I'll also make sure to look into your hint on the alignment. 1 Link to comment Share on other sites More sharing options...

## Recommended Posts

## Create an account or sign in to comment

You need to be a member in order to leave a comment

## Create an account

Sign up for a new account in our community. It's easy!

Register a new account## Sign in

Already have an account? Sign in here.

Sign In Now