Math Is Fun Forum

George,Y

Member

Registered: 2006-03-12

Posts: 1,379

Tricky Combination Equality

C[sub]n+1[/sub][sup]m+1[/sup]

=

C[sub]n[/sub][sup]n[/sup]+C[sub]n[/sub][sup]n+1[/sup]+...+C[sub]n[/sub][sup]m[/sup]

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X'(y-Xβ)=0

George,Y

Member

Registered: 2006-03-12

Posts: 1,379

Re: Tricky Combination Equality

5 choose 3
= 2 choose 2 + 3 choose 2 + 4 choose 2

10=1+3+6

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X'(y-Xβ)=0

Bob

Administrator

Registered: 2010-06-20

Posts: 10,627

Re: Tricky Combination Equality

hi George, Y,

Interesting.  Are you offering this as an interesting bit of maths or were you hoping for a proof?

Bob

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Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Tricky Combination Equality

Hi George,Y;

You might consider an n x n grid called a lattice and doing a block walk on it. That is the usual way identities like that are proved.

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In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

George,Y

Member

Registered: 2006-03-12

Posts: 1,379

Re: Tricky Combination Equality

Hi, I know a proof through Pascal's triangle, but can anyone latex the combination sign?

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X'(y-Xβ)=0

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Tricky Combination Equality

Hi George,Y;

There are two ways.

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In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

George,Y

Member

Registered: 2006-03-12

Posts: 1,379

Re: Tricky Combination Equality

First, a review on Pascal's Triangle

The intuition is n already chosen from m and then add 1 to m and choose n+1 instead.  the added 1 to pool of m can be either selected with n already chosen or neglected. In the former case m choose n, the latter case m choose n+1. The two way conclude all the possible ways to choose n+1 from m+1.

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X'(y-Xβ)=0

George,Y

Member

Registered: 2006-03-12

Posts: 1,379

Re: Tricky Combination Equality

The idea need proof is

Proof:

......

&

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X'(y-Xβ)=0