Difference between revisions of "Combination"
IntrepidMath (talk | contribs) |
|||
Line 18: | Line 18: | ||
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>. | ||
+ | |||
=== See also === | === See also === |
Revision as of 14:14, 18 June 2006
Contents
[hide]Definition
The number of combinations of objects from a set of objects is the number of ways the objects can be arranged with regard to order.
Notation
The common forms of denoting the number of combinations of objects from a set of objects is:
Formula
Derivation
Consider the set of letters A, B, and C. There are 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 objects from elements , there are more ways to permute them than to choose them. We have , or .