Difference between revisions of "Combination"

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<math>\sum_{x=0}^{n} \binom{n}{x} = 2^n</math>
 
<math>\sum_{x=0}^{n} \binom{n}{x} = 2^n</math>
  
 +
One of the many proofs is by first inserting <math>a = b = 1</math> in to the [[binomial theorem]]. Because the combinations are the coefficients of <math>2^n</math>, and a and b cancel out because they are 1, the sum is <math>2^n</math>.
  
 
== Examples ==
 
== Examples ==

Revision as of 19:17, 3 November 2007

This is an AoPSWiki Word of the Week for Nov 1-7

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:

  • $\binom{n}{r}$
  • ${C}(n,r)$
  • $\,_{n} C_{r}$
  • $C_n^{r}$

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

Other Formulas

$\binom{n-1}{r-1}+\binom{n-1}{r}=\binom{n}{r}$

$\sum_{x=0}^{n} \binom{n}{x} = 2^n$

One of the many proofs is by first inserting $a = b = 1$ in to the binomial theorem. Because the combinations are the coefficients of $2^n$, and a and b cancel out because they are 1, the sum is $2^n$.

Examples

Links

See also

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