Difference between revisions of "Combination"

(Examples: added 2000II/5)
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=== Definition ===
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== Introduction ==
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A '''combination''' is a way of choosing <math>r</math> objects from a set of <math>n</math> where the order in which the objects are chosen is irrelevant.  We are generally concerned with finding the number of combinations of size <math>r</math> from an original set of size <math>n</math>
  
The number of '''combinations''' of <math>{r}</math> objects from a set of <math>{n}</math> objects is the number of ways the <math>{r}</math> objects can be arranged ''with regard to order''.
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== Notation ==
 
 
=== Notation ===
 
  
 
The common forms of denoting the number of combinations of <math>{r}</math> objects from a set of <math>{n}</math> objects is:
 
The common forms of denoting the number of combinations of <math>{r}</math> objects from a set of <math>{n}</math> objects is:
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* <math>{C}(n,r)</math>
 
* <math>{C}(n,r)</math>
 
* <math>\,_{n} C_{r}</math>
 
* <math>\,_{n} C_{r}</math>
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* <math> C^n_{r} </math>
  
=== Formula ===
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== Formula ==
  
 
<math>{{n}\choose {r}} = \frac {n!} {r!(n-r)!}</math>
 
<math>{{n}\choose {r}} = \frac {n!} {r!(n-r)!}</math>
  
=== Derivation ===
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== Derivation ==
  
 
Consider the set of letters A, B, and C.  There are <math>3!</math> different [[permutations]] of those letters. Since order doesn't matter with combinations, there is only one combination of those three.  In general, since for every permutation of <math>{r}</math> objects from <math>{n}</math> elements <math>P(n,r)</math>, there are <math>{r}!</math> more ways to permute them than to choose them.  We have <math>{r}!{C}({n},{r})=P(n,r)</math>, or <math>{{n}\choose {r}} = \frac {n!} {r!(n-r)!}</math>.
 
Consider the set of letters A, B, and C.  There are <math>3!</math> different [[permutations]] of those letters. Since order doesn't matter with combinations, there is only one combination of those three.  In general, since for every permutation of <math>{r}</math> objects from <math>{n}</math> elements <math>P(n,r)</math>, there are <math>{r}!</math> more ways to permute them than to choose them.  We have <math>{r}!{C}({n},{r})=P(n,r)</math>, or <math>{{n}\choose {r}} = \frac {n!} {r!(n-r)!}</math>.
  
  
=== Examples ===
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== Examples ==
  
 
* [http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=45&year=2005&p=368223 AIME 2005II/1]
 
* [http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=45&year=2005&p=368223 AIME 2005II/1]
 
* [http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=45&year=2000&p=385889 AIME 2000II/5]
 
* [http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=45&year=2000&p=385889 AIME 2000II/5]
  
=== See also ===
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== See also ==
  
 
* [[Combinatorics]]
 
* [[Combinatorics]]
 
* [[Permutations]]
 
* [[Permutations]]
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* [[Pascal's Triangle]]

Revision as of 22:50, 20 June 2006

Introduction

A combination is a way of choosing $r$ objects from a set of $n$ where the order in which the objects are chosen is irrelevant. We are generally concerned with finding the number of combinations of size $r$ from an original set of size $n$

Notation

The common forms of denoting the number of combinations of ${r}$ objects from a set of ${n}$ objects is:

  • ${n}\choose {r}$
  • ${C}(n,r)$
  • $\,_{n} C_{r}$
  • $C^n_{r}$

Formula

${{n}\choose {r}} = \frac {n!} {r!(n-r)!}$

Derivation

Consider the set of letters A, B, and C. There are $3!$ different permutations of those letters. Since order doesn't matter with combinations, there is only one combination of those three. In general, since for every permutation of ${r}$ objects from ${n}$ elements $P(n,r)$, there are ${r}!$ more ways to permute them than to choose them. We have ${r}!{C}({n},{r})=P(n,r)$, or ${{n}\choose {r}} = \frac {n!} {r!(n-r)!}$.


Examples

See also