Difference between revisions of "Combinatorial identity"

Revision as of 13:38, 3 August 2010

Vandermonde's Identity

Vandermonde's Identity states that $\sum_{k=0}^r\binom mk\binom n{r-k}=\binom{m+n}r$ , which can be proven combinatorially by noting that any combination of $r$ objects from a group of $m+n$ objects must have some $0\le k\le r$ objects from group $m$ and the remaining from group $n$ .

Another Identity

$\sum_{i=0}^k \binom{k}{i}^2=\binom{2k}{k}$

Hat Proof

We have $2k$ different hats. We split them into two groups, each with k hats: then we choose $i$ hats from the first group and $k-i$ hats from the second group. This may be done in $\binom{k}{i}^2$ ways. Evidently, to generate all possible choices of $k$ hats from the $2k$ hats, we must choose $i=0,1,\cdots,k$ hats from the first $k$ and the remaining $k-i$ hats from the second $k$ ; the sum over all such $i$ is the number of ways of choosing $k$ hats from $2k$ . Therefore $\sum_{i=0}^k \binom{k}{i}^2=\binom{2k}{k}$ , as desired.

@@ Line 1: / Line 1: @@
-==Hockey-Stick Identity==
-For <math>n,r\in\mathbb{N}, n>r,\sum^n_{i=r}{i\choose r}={n+1\choose r+1}</math>.
-<asy>
-int chew(int n,int r){
- int res=1;
- for(int i=0;i<r;++i){
-  res=quotient(res*(n-i),i+1);
-  }
- return res;
- }
-for(int n=0;n<9;++n){
- for(int i=0;i<=n;++i){
-  if((i==2 && n<8)||(i==3 && n==8)){
-   if(n==8){label(string(chew(n,i)),(11+n/2-i,-n),p=red+2.5);}
-   else{label(string(chew(n,i)),(11+n/2-i,-n),p=blue+2);}
-   }
-  else{
-   label(string(chew(n,i)),(11+n/2-i,-n));
-   }
-  }
- }
-</asy>
-This identity is known as the ''hockey-stick'' identity because, on Pascal's triangle, when the addends represented in the summation and the sum itself are highlighted, a hockey-stick shape is revealed.
-===Proof===
-'''Inductive Proof'''
-This identity can be proven by induction on <math>n</math>.
-<u>Base Case</u>
-Let <math>n=r</math>.
-<math>\sum^n_{i=r}{i\choose r}=\sum^r_{i=r}{i\choose r}={r\choose r}=1={r+1\choose r+1}</math>.
-<u>Inductive Step</u>
-Suppose, for some <math>k\in\mathbb{N}, k>r</math>, <math>\sum^k_{i=r}{i\choose r}={k+1\choose r+1}</math>.
-Then <math>\sum^{k+1}_{i=r}{i\choose r}=\left(\sum^k_{i=r}{i\choose r}\right)+{k+1\choose r}={k+1\choose r+1}+{k+1\choose r}={k+2\choose r+1}</math>.
-'''Algebraic Proof'''
-It can also be proven algebraically with [[Pascal's Identity]], <math>{n \choose k}={n-1\choose k-1}+{n-1\choose k}</math>.
-Note that
-<math>{r \choose r}+{r+1 \choose r}+{r+2 \choose r}+\cdots+{r+a \choose r}</math>
-<math>={r+1 \choose r+1}+{r+1 \choose r}+{r+2 \choose r}+\cdots+{r+a \choose r}</math>
-<math>={r+2 \choose r+1}+{r+2 \choose r}+\cdots+{r+a \choose r}=\cdots={r+a \choose r-1}+{r+a \choose r}={r+a+1 \choose r+1}</math>, which is equivalent to the desired result.
-'''Combinatorial Proof'''
-Imagine that we are distributing <math>n</math> indistinguishable candies to <math>k</math> distinguishable children. By a direct application of Balls and Urns, there are <math>{n+k-1\choose k-1}</math> ways to do this. Alternatively, we can first give <math>0\le i\le n</math> candies to the oldest child so that we are essentially giving <math>n-i</math> candies to <math>k-1</math> kids and again, with Balls and Urns, <math>{n+k-1\choose k-1}=\sum_{i=0}^n{n+k-2-i\choose k-2}</math>, which simplifies to the desired result.
 ==Vandermonde's Identity==
 Vandermonde's Identity states that <math>\sum_{k=0}^r\binom mk\binom n{r-k}=\binom{m+n}r</math>, which can be proven combinatorially by noting that any combination of <math>r</math> objects from a group of <math>m+n</math> objects must have some <math>0\le k\le r</math> objects from group <math>m</math> and the remaining from group <math>n</math>.

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Difference between revisions of "Combinatorial identity"

Revision as of 13:38, 3 August 2010

Contents

Vandermonde's Identity

Another Identity

Hat Proof

Examples

See also