Difference between revisions of "Combination"

Revision as of 22:32, 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 ${{n}\choose {r}} = \frac {n!} {r!(n-r)!}$ .

Formulas/Identities

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

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

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

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 $2^n$ .

$\sum_{i=0}^{k}\binom{m}{i}\binom{n}{k-i}=\binom{m+n}{k}$

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

We can prove this by putting the combinations in their algebraic form. $\binom{n}{n-r}=\frac{n!}{(n-r)!(n-(n-r)!}$ . As we can see, $(n-(n-r)!=(n-n+r)=r!$ . By the commutative property, $\frac{n!}{(n-r)!r!}=\frac{n!}{r!(n-r)!}$ . Because $\frac{n!}{r!(n-r)!}=\binom{n}{r}$ , by the transitive property, we can conclude that this is true for all non-negative integers n and r where n is greater than or equal to r. Another reason this is true is because that choosing what you don't what is the same as choosing what you do want, because by choosing what you don't want, you imply that you choose the rest.

Examples

Links

[1]

@@ Line 31: / Line 31: @@
 * <math>\sum_{i=0}^{k}\binom{m}{i}\binom{n}{k-i}=\binom{m+n}{k}</math>
+* <math>\binom{n}{r}=\binom{n}{n-r}</math>
+We can prove this by putting the combinations in their algebraic form.
+<math>\binom{n}{n-r}=\frac{n!}{(n-r)!(n-(n-r)!}</math>. As we can see, <math>(n-(n-r)!=(n-n+r)=r!</math>. By the [[commutative property]], <math>\frac{n!}{(n-r)!r!}=\frac{n!}{r!(n-r)!}</math>. Because <math>\frac{n!}{r!(n-r)!}=\binom{n}{r}</math>, by the [[transitive property]], we can conclude that this is true for all non-negative integers n and r where n is greater than or equal to r. Another reason this is true is because that choosing what you don't what is the same as choosing what you do want, because by choosing what you don't want, you imply that you choose the rest.
 == Examples ==

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