Difference between revisions of "Combinatorial identity"

Revision as of 11:29, 10 March 2009

Hockey-Stick Identity

For $n,r\in\mathbb{N}, n>r,\sum^n_{i=r}{i\choose r}={n+1\choose r+1}$ .

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.

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Proof

Inductive Proof

This identity can be proven by induction on $n$ .

Base Case Let $n=r$ .

$\sum^n_{i=r}{i\choose r}=\sum^r_{i=r}{i\choose r}={r\choose r}=1={r+1\choose r+1}$ .

Inductive Step Suppose, for some $k\in\mathbb{N}, k>r$ , $\sum^k_{i=r}{i\choose r}={k+1\choose r+1}$ . Then $\sum^{k+1}_{i=r}{i\choose r}=\left(\sum^k_{i=r}{i\choose r}\right)+{k+1\choose r}={k+1\choose r+1}+{k+1\choose r}={k+2\choose r+1}$ .

Algebraic Proof

It can also be proven algebraically with Pascal's Identity, ${n \choose k}={n-1\choose k-1}+{n-1\choose k}$ . Note that

${r \choose r}+{r+1 \choose r}+{r+2 \choose r}+\cdots+{r+a \choose r}$ $={r+1 \choose r+1}+{r+1 \choose r}+{r+2 \choose r}+\cdots+{r+a \choose r}$ $={r+2 \choose r+1}+{r+2 \choose r}+\cdots+{r+a \choose r}=\cdots={r+a \choose r-1}+{r+a \choose r}={r+a+1 \choose r+1}$ , which is equivalent to the desired result.

Combinatorial Proof

Imagine that we are distributing $n$ indistinguishable candies to $k$ distinguishable children. By a direct application of Balls and Urns, there are ${n+k-1\choose k-1}$ ways to do this. Alternatively, we can first give $0\le i\le n$ candies to the oldest child so that we are essentially giving $n-i$ candies to $k-1$ kids and again, with Balls and Urns, ${n+k-1\choose k-1}=\sum_{i=0}^n{n+k-2-i\choose k-2}$ , which simplifies to the desired result.

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>.
@@ Line 37: / Line 36: @@
 ==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>.
+==Another Identity==
+\[
+\sum_{i=0}^k \binom{k}{i}^2=\binom{2k}{k}
+\]
+===Hat Proof===
+We have <math>2k</math> different hats. We split them into two groups, each with k hats: then we choose <math>i</math> hats from the first group and <math>k-i</math> hats from the second group. This may be done in <math>\binom{k}{i}^2</math> ways. Evidently, to generate all possible choices of <math>k</math> hats from the <math>2k</math> hats, we must choose <math>i=0,1,\cdots,k</math> hats from the first <math>k</math> and the remaining <math>k-i</math> hats from the second <math>k</math>; the sum over all such <math>i</math> is the number of ways of choosing <math>k</math> hats from <math>2k</math>. Therefore  <math>\sum_{i=0}^k \binom{k}{i}^2=\binom{2k}{k}</math>, as desired.
 == Examples ==

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