Hello everyone

 

In math class, we solved the next problem:

 

How many numbers, composed out of 5 different digits, varying from 0 to 6, can be formed?

So we concluded that the number of possibilities = [math]6\cdot 6\cdot 5\cdot 4\cdot 3[/math], excluding 0 as first digit.

 

Conclusion: [math]N=6\cdot V^4_6[/math].

 

Then, suddenly, something came in my head; as we were working with variation formulas, I wanted to put the number of possibilities in a formula, solely using variation formulas; the first thing that came in my head was:

[math]V^5_7-V^4_6[/math]

My math teacher did some thinking and accepted my resolution (she said it was also a good solution), my reasoning to make her tell me I'm right: "every possible numbers, also commencing with 0 - the numbers commencing with 0 (excluding 0, but keeping in mind that 0 is already 'in' the number, there are 4 digits to be picked out of 6 possible digits)"

 

So, I was right. Now, I wanted to see if this was only for this problem, or for every problem. So I set up a general formula:

[math]n\cdot V^{n-2}_n = V^{n-1}_{n+1}-V^{n-2}_n[/math]

 

I have proven this, by starting with this equation and finally getting the equation [math]n+1=n+1[/math], which is true.

 

Now I was wondering, is this formula utile? Can it be used for something? For what?

 

Thanks

 

Function

For those who are eager to see the 'proof':

 

[math]n\cdot V^{n-2}_{n}=V^{n-1}_{n+1}-V^{n-2}_{n}[/math]

 

[math]\Leftrightarrow n\cdot V^{n-2}_{n}=V^{n-2}_{n}\left(\frac{V^{n-1}_{n+1}}{V^{n-2}_{n}}-1\right)[/math]

 

[math]\Leftrightarrow n+1=\frac{V^{n-1}_{n+1}}{V^{n-2}_{n}}[/math]

 

[math]\Leftrightarrow n+1=\frac{(n+1)n(n-1)(n-2)\cdots (n+1-n+1+1)}{n(n-1)(n-2)\cdots (n-n+2+1)}[/math]

 

[math]\Leftrightarrow n+1=n+1[/math]

 

True

 

[math]\Leftrightarrow n\cdot V^{n-2}_{n}=V^{n-1}_{n+1}-V^{n-2}_{n}[/math]

 

Q.E.D.

 

(P.S. is there a name for proofs which start with the theorem?)

Edited January 24, 201411 yr by Function

EDIT: A bit more general:

 

[math]n\cdot V^{n-m}_n = V^{n-m+1}_{n+1}-V^{n-m}_n[/math]

 

[math]\Leftrightarrow n\cdot V^{n-m}_n=V^{n-m}_n\left(\frac{V^{n-m+1}_{n+1}}{V^{n-m}_n}-1\right)[/math]

 

[math]\Leftrightarrow n+1=\frac{(n+1)n(n-1)\cdots (n-n+m-1+1)}{n(n-1)\cdots (n-n+m)}=\frac{(n+1)m}{m}=n+1[/math]

 

True

 

[math]\Leftrightarrow n\cdot V^{n-m}_n = V^{n-m+1}_{n+1}-V^{n-m}_n[/math]

 

Q.E.D.

 

ALTERNATIVE PROOF:

 

[math]n\cdot V^{n-m}_n = V^{n-m+1}_{n+1}-V^{n-m}_n[/math]

 

[math]\Leftrightarrow n\cdot\frac{n!}{m!}=\frac{(n+1)!}{m!}-\frac{n!}{m!}[/math]

 

[math]\Leftrightarrow n^2(n-1)!=(n+1)!-n![/math]

 

[math]\Leftrightarrow n^2(n-1)!=(n+1)n(n-1)\cdots 2\cdot 1 - n(n-1)\cdots 2\cdot 1[/math]

 

[math]\Leftrightarrow n^2(n-1)!=n!\cdot n[/math]

 

[math]\Leftrightarrow n^2(n-1)(n-2)\cdots 2\cdot 1 = n\cdot n(n-1)(n-2)\cdots 2\cdot 1[/math]

 

[math]\Leftrightarrow n^2(n-1)! = n^2(n-1)![/math]

 

True

 

[math]n\cdot V^{n-m}_n = V^{n-m+1}_{n+1}-V^{n-m}_n[/math]

Q.E.D.

Edited February 2, 201411 yr by Function