Can someone tell me how this works?

>>8799399
Sure. Few ways to look at it. That equation is wrong by the way. The first term on the left has m!/m!. Probably the easiest is to use binomials or pascals triangle. However you could just simplify both sides instead.

[eqn] (x+y)^n = \sum_{k=0}^n \frac{n!}{k!(n-k)!}x^k y^(n-k) [/eqn]
Multiply both sides by [math] x+y [/math] to get
[eqn] (x+y)^(n+1) = \sum_{k=0}^{n+1} \Big(\frac{n!}{k! (n-k)!}+\frac{n!}{(k-1)! (n-k+1)!}\Big)x^k y^(n-k+1) [/eqn]

Equation equal coefficients in this equation with the coefficients in the first equation after replacing [math] n [/math] by [math] n+1 [/math]

Paraphrasing a VERY smart person.
"No one understands why math works. They just get used to it and take that for understanding"

>>8799432
Obviously I accidentally powered just the parenthese and not the whole term in the [math] n-k, n-k+1, \text{and} n+1 [/math]

There's another clever way to look at it using combinatorics.

Suppose you have n+1 objects and want to know how many ways you can pick out k objects.

First set one object aside. Decide you won't use that object, and calculate how many ways you can pick k objects out of n objects. Now also decide you WILL use that object. Calculate how many ways you can pick k-1 objects out of n objects.

Obviously if you add these you'll just get how many ways you can pick out k objects out of n+1 objects.

tada, two ways to prove the same thing.

>>8799448
>>8799432
>>8799399
make sense OP?

>>8799399
[math]\frac{m!}{n!(m-n)!}+\frac{m!}{(m-n+1)!(n-1)!}=[/math]
[math]\frac{m!(m-n+1)}{n!(m-n+1)!}+\frac{m!n}{(m-n+1)!n!}=[/math]
[math]\frac{m!}{n!(m-n+1)!}(m-n+1+n)=\frac{m!(m+1)}{n!(m-n+1)!}[/math]

File: Snapchat-2025479636.jpg (387KB, 1080x1920px) Image search: [Google]

387KB, 1080x1920px

>>8799448
this is how the problem is worked out in my book. It is in Serge Lang's Basic Mathematics

>>8799458
As I said in the first proof it's probably easiest just to simplify it. However proving it using the binomial expansion or this combinatorics proof seems to give a bit more insight.

>>8799458
I don't understand how the second numerator turns into m!n

>>8799465
Because 1/(n-1)!=n/n!

>>8799468
thank you so much

>>8799468
>>8799460
>>8799456
>>8799457
>>8799448
>>8799441
>>8799432
Thank you, all. Very helpful