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| {{Redirect|Combin|the mountain massif|Grand Combin}}
| | Building backlinks is very important when trying to get your web site a great ranking. Although maybe not the sole significant aspect of SEO, it nevertheless has loads of effect and as such should be attentively attacked. Several sites attempt to con search motors in to boosting their position by making backlinks through loads of low quality sites. The largest research engines changed their algorithms therefore the quality of the backlinks become more significant as opposed to quantity, due to the negative effect this black hat method had on the attribute so search engine results. Sites that had inferior quality backlinks that arrived about overnight were punished and the ones that had cultivated a steady rate of large PR links were promoted.<br><br>In SEO backlinks are important because they help to boost traffic amounts. Besides search engine outcomes, online users will frequently uncover themselves researching new sites simply as a result of a hyperlink they discovered on yet another site. Most often the best hyperlink is whatever comes from a site that offers that will be highly relevant to the on-line users. When these consumers make visits to the internet site on another end-of the hyperlink, they are able to then more readily be persuaded by the quality articles and goods provided to create a buy. With increased traffic being aimed to the business website by backlinks, the more chances there are to boost sales amounts and guarantee a selling. As such, backlinks make it possible for a business to make more income and profit.<br><br>Another way where backlinks have a positive effect on SEO is through fostering the number of visits from internet search engine crawlers. These crawlers will most likely index extremely rated sites frequently as a result of their several visitants along with the quality content. They follow them to discover if they guide to similarly quality content and catalog it and the other characteristics of the site, when they happen upon backlinks. With improved indexing, quality articles plus a properly designed website, their ranking can be seen by a business improved radically. It's crucial nevertheless to ensure your the website along with content are appropriately done up to impress both the on-line guests and search engine robots. With quality articles, indexing, more visitors and upgrades, a website may anticipate to rise to the top most pages of the internet search motor outcomes. This is an achievement that's possible for both large and smaller businesses when proper SEARCH ENGINE OPTIMIZATION TECHNIQUES are employed consistently. As you can see on [http://www.scrapethatrun1112.org Scrape IT 007]. |
| {{Other uses}}
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| In [[mathematics]], a '''combination''' is a way of selecting several things out of a larger group, where (unlike [[permutations]]) order does not matter. In smaller cases it is possible to count the number of combinations. For example given three fruits, say an apple, an orange and a pear, there are three combinations of two that can be drawn from this set: an apple and a pear; an apple and an orange; or a pear and an orange.
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| More formally, a ''k''-'''combination''' of a [[Set (mathematics)|set]] ''S'' is a subset of ''k'' distinct elements of ''S''. If the set has ''n'' elements the number of ''k''-combinations is equal to the [[binomial coefficient]]
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| :<math> \binom nk = \frac{n(n-1)\ldots(n-k+1)}{k(k-1)\dots1},</math>
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| which can be written using [[factorial]]s as <math>\frac{n!}{k!(n-k)!}</math> whenever <math>k\leq n</math>, and which is zero when <math>k>n</math>.
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| The set of all ''k''-combinations of a set ''S'' is sometimes denoted by <math>\binom Sk\,</math>.
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| Combinations refer to the combination of ''n'' things taken ''k'' at a time without repetition. To refer to combinations in which repetition is allowed, the terms ''k''-selection,<ref>{{harvnb|Ryser|1963|loc=p. 7}} also referred to as an ''unordered selection''.</ref> ''k''-multiset,<ref>{{harvnb|Mazur|2010|loc=p. 10}}</ref> or ''k''-combination with repetition are often used.<ref>When the term ''combination'' is used to refer to either situation (as in {{harv|Brualdi|2010}}) care must be taken to clarify whether sets or multisets are being discussed.</ref> If, in the above example, it was possible to have two of any one kind of fruit there would be 3 more 2-selections: one with two apples, one with two oranges, and one with two pears.
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| With large sets, it becomes necessary to use more sophisticated mathematics to find the number of combinations. For example, a [[Hand (poker)|poker hand]] can be described as a 5-combination (''k'' = 5) of cards from a 52 card deck (''n'' = 52). The 5 cards of the hand are all distinct, and the order of cards in the hand does not matter. There are 2,598,960 such combinations, and the chance of drawing any one hand at random is 1 / 2,598,960.
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| == Number of ''k''-combinations ==
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| [[File:Combinations without repetition; 5 choose 3.svg|thumb|3-element subsets of a 5-element set]]
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| {{main|Binomial coefficient}}
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| The [[Permutations_and_combinations#case_in|number of ''k''-combinations]] from a given set ''S'' of ''n'' elements is often denoted in elementary combinatorics texts by ''C''(''n'', ''k''), or by a variation such as <math>C^n_k</math>, <math>{}_nC_k</math>, <math>{}^nC_k</math> or even <math>C_n^k</math> (the latter form was standard in French, Russian, Chinese<ref>{{cite book |title = High School Textbook for full-time student (Required) Mathematics Book II B| edition=2nd | location = China|language = Chinese | year= 2006| publisher = People's Education Press| pages = 107-116 | month = June| isbn = 978-7-107-19616-4 }}</ref> and Polish texts{{citation needed|date=April 2012}}). The same number however occurs in many other mathematical contexts, where it is denoted by <math>\tbinom nk</math> (often read as "n choose k"); notably it occurs as a coefficient in the [[binomial formula]], hence its name '''binomial coefficient'''. One can define <math>\tbinom nk</math> for all natural numbers ''k'' at once by the relation
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| :<math>\textstyle(1+X)^n=\sum_{k\geq0}\binom nk X^k,</math>
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| from which it is clear that <math>\tbinom n0=\tbinom nn=1</math> and <math>\tbinom nk=0</math> for ''k'' > ''n''. To see that these coefficients count ''k''-combinations from ''S'', one can first consider a collection of ''n'' distinct variables ''X''<sub>''s''</sub> labeled by the elements ''s'' of ''S'', and expand the [[Multiplication|product]] over all elements of ''S'':
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| :<math>\textstyle\prod_{s\in S}(1+X_s);</math>
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| it has 2<sup>''n''</sup> distinct terms corresponding to all the subsets of ''S'', each subset giving the product of the corresponding variables ''X''<sub>''s''</sub>. Now setting all of the ''X''<sub>''s''</sub> equal to the unlabeled variable ''X'', so that the product becomes {{nowrap|(1 + ''X'')<sup>''n''</sup>}}, the term for each ''k''-combination from ''S'' becomes ''X''<sup>''k''</sup>, so that the coefficient of that power in the result equals the number of such ''k''-combinations.
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| Binomial coefficients can be computed explicitly in various ways. To get all of them for the expansions up to {{nowrap|(1 + ''X'')<sup>''n''</sup>}}, one can use (in addition to the basic cases already given) the recursion relation
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| :<math>\binom nk=\binom{n-1}{k-1}+\binom{n-1}k,\text{ for }0<k<n,</math>
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| which follows from {{nowrap|(1 + ''X'')<sup>''n''</sup> }}={{nowrap| (1 + ''X'')<sup>''n'' − 1</sup>(1 + ''X'')}}; this leads to the construction of [[Pascal's triangle]].
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| For determining an individual binomial coefficient, it is more practical to use the formula
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| :<math>\binom nk = \frac{n(n-1)(n-2)\cdots(n-k+1)}{k!}.</math>
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| The [[numerator]] gives the number of [[Permutations_and_combinations#case_i|''k''-permutations]] of ''n'', i.e., of sequences of ''k'' distinct elements of ''S'', while the [[denominator]] gives the number of such ''k''-permutations that give the same ''k''-combination when the order is ignored.
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| When ''k'' exceeds ''n''/2, the above formula contains factors common to the numerator and the denominator, and canceling them out gives the relation
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| :<math> \binom nk = \binom n{n-k},\text{ for }0 \le k \le n.</math>
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| This expresses a symmetry that is evident from the binomial formula, and can also be understood in terms of ''k''-combinations by taking the [[complement (set theory)|complement]] of such a combination, which is an {{nowrap|(''n'' − ''k'')}}-combination.
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| Finally there is a formula which exhibits this symmetry directly, and has the merit of being easy to remember:
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| :<math> \binom nk = \frac{n!}{k!(n-k)!},</math>
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| where ''n''<nowiki>!</nowiki> denotes the [[factorial]] of ''n''. It is obtained from the previous formula by multiplying denominator and numerator by {{nowrap|(''n'' − ''k'')}}!, so it is certainly inferior as a method of computation to that formula.
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| The last formula can be understood directly, by considering the ''n''<nowiki>!</nowiki> permutations of all the elements of ''S''. Each such permutation gives a ''k''-combination by selecting its first ''k'' elements. There are many duplicate selections: any combined permutation of the first ''k'' elements among each other, and of the final (''n'' − ''k'') elements among each other produces the same combination; this explains the division in the formula.
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| From the above formulas follow relations between adjacent numbers in Pascal's triangle in all three directions:
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| :<math> \binom nk = \binom n{k-1} \frac {n-k+1}k,\text{ for }k>0 </math>,
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| :<math> \binom nk = \binom {n-1}k \frac n{n-k},\text{ for }{k<n} </math>,
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| :<math> \binom nk = \binom {n-1}{k-1} \frac nk,\text{ for }n,k>0 </math>.
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| Together with the basic cases <math>\tbinom n0=1=\tbinom nn</math>, these allow successive computation of respectively all numbers of combinations from the same set (a row in Pascal's triangle), of ''k''-combinations of sets of growing sizes, and of combinations with a complement of fixed size {{nowrap|''n'' − ''k''}}.
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| === Example of counting combinations ===
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| As a concrete example, one can compute the number of five-card hands possible from a standard fifty-two card deck as:<ref>{{harvnb|Mazur|2010|loc=p. 21}}</ref>
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| :<math> {52 \choose 5} = \frac{52\times51\times50\times49\times48}{5\times4\times3\times2\times1} = \frac{311,875,200}{120} =
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| 2,598,960.</math>
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| Alternatively one may use the formula in terms of factorials and cancel the factors in the numerator against parts of the factors in the denominator, after which only multiplication of the remaining factors is required:
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| :<math> \begin{alignat}{2}{52 \choose 5} &= \frac{52!}{5!47!} \\
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| &= \frac{52\times51\times50\times49\times48\times\cancel{47!}}{5\times4\times3\times2\times\cancel{1}\times\cancel{47!}} \\
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| &= \frac{52\times51\times50\times49\times48}{5\times4\times3\times2} \\
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| &= \frac{(26\times\cancel{2})\times(17\times\cancel{3})\times(10\times\cancel{5})\times49\times(12\times\cancel{4})}{\cancel{5}\times\cancel{4}\times\cancel{3}\times\cancel{2}} \\
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| &= {26\times17\times10\times49\times12} \\&= 2,598,960.\end{alignat}</math>
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| Another alternative computation, equivalent to the first, is based on writing
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| :<math> {n \choose k} = \frac { ( n - 0 ) }1 \times \frac { ( n - 1 ) }2 \times \frac { ( n - 2 ) }3 \times \cdots \times \frac { ( n - (k - 1) ) }k,</math>
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| which gives
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| :<math> {52 \choose 5} = \frac{52}1 \times \frac{51}2 \times \frac{50}3 \times \frac{49}4 \times \frac{48}5 = 2,598,960.</math>
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| When evaluated in the following order, {{math|52 ÷ 1 × 51 ÷ 2 × 50 ÷ 3 × 49 ÷ 4 × 48 ÷ 5}}, this can be computed using only integer arithmetic. The reason is that when each division occurs, the intermediate result that is produced is itself a binomial coefficient, so no remainders ever occur.{{cn|date=December 2013}}
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| Using the symmetric formula in terms of factorials without performing simplifications gives a rather extensive calculation:
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| :<math>
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| \begin{align}
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| {52 \choose 5} &= \frac{n!}{k!(n-k)!} = \frac{52!}{5!(52-5)!} = \frac{52!}{5!47!} \\
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| &= \tfrac{80,658,175,170,943,878,571,660,636,856,403,766,975,289,505,440,883,277,824,000,000,000,000}{120\times258,623,241,511,168,180,642,964,355,153,611,979,969,197,632,389,120,000,000,000} \\
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| &= 2,598,960.
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| \end{align}</math>
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| === Enumerating ''k''-combinations ===
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| One can [[enumerate]] all ''k''-combinations of a given set {{mvar|''S''}} of ''n'' elements in some fixed order, which establishes a [[bijection]] from an interval of <math>\tbinom nk</math> integers with the set of those ''k''-combinations. Assuming {{mvar|''S''}} is itself ordered, for instance {{math|''S'' {{=}} {1,2, ...,''n''}}}, there are two natural possibilities for ordering its ''k''-combinations: by comparing their smallest elements first (as in the illustrations above) or by comparing their largest elements first. The latter option has the advantage that adding a new largest element to {{mvar|''S''}} will not change the initial part of the enumeration, but just add the new ''k''-combinations of the larger set after the previous ones. Repeating this process, the enumeration can be extended indefinitely with ''k''-combinations of ever larger sets. If moreover the intervals of the integers are taken to start at 0, then the ''k''-combination at a given place ''i'' in the enumeration can be computed easily from ''i'', and the bijection so obtained is known as the [[combinatorial number system]]. It is also known as "rank"/"ranking" and "unranking" in computational mathematics.<ref>http://www.site.uottawa.ca/~lucia/courses/5165-09/GenCombObj.pdf</ref><ref>http://www.sagemath.org/doc/reference/sage/combinat/subset.html</ref>
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| There are many ways to enumerate ''k'' combinations. One way is to visit all the binary numbers less than <math>2^n</math>. Chose those numbers having ''k'' nonzero bits. The positions of these 1 bits in such a number is a specific ''k''-combination of the set {1,...,''n''}.<ref>http://rosettacode.org/wiki/Combinations</ref>
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| == Number of combinations with repetition ==
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| {{See also|Multiset coefficient}}
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| [[File:Combinations with repetition; 5 multichoose 3.svg|thumb|370px|[[Bijection]] between 3-subsets of a 7-set (left)<br>and 3-multisets with elements from a 5-set (right)<br>So this illustrates that <math>\textstyle {7 \choose 3} = \left(\!\!{5 \choose 3}\!\!\right)</math>.]]
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| A ''k''-'''combination with repetitions''', or ''k''-'''multicombination''', or '''[[multiset|multisubset]]''' of size ''k'' from a set ''S'' is given by a sequence of ''k'' not necessarily distinct elements of ''S'', where order is not taken into account: two sequences of which one can be obtained from the other by permuting the terms define the same multiset. In other words, the number of ways to sample ''k'' elements from a set of ''n'' elements allowing for duplicates (i.e., with replacement) but disregarding different orderings (e.g. {2,1,2} = {1,2,2}). Associate an index to each element of ''S'' and think of the elements of ''S'' as ''types'' of objects, then we can let <math>x_i</math> denote the number of elements of type ''i'' in a multisubset. The number of multisubsets of size ''k'' is then the number of nonnegative integer solutions of the [[Diophantine equation]]:<ref>{{harvnb|Brualdi|2010|loc=p. 52}}</ref>
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| :<math>x_1 + x_2 + \ldots + x_n = k.</math>
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| If ''S'' has ''n'' elements, the number of such ''k''-multisubsets is denoted by,
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| ::<math>\left(\!\!\!\binom{n}{k}\!\!\!\right),</math>
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| a notation that is analogous to the [[binomial coefficient]] which counts ''k''-subsets. This expression, ''n'' multichoose ''k'',<ref>{{harvnb|Benjamin|Quinn|2003|loc=p. 70}}</ref> is also given by a binomial coefficient: | |
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| :<math>\left(\!\!\!\binom{n}{k}\!\!\!\right)=\binom{n+k-1}{k}.</math>
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| This relationship can be easily seen using a representation known as [[Stars and bars (combinatorics)|stars and bars]]. A solution of the above Diophantine equation can be represented by <math>x_1</math> ''stars'', a separator (a ''bar''), then <math>x_2</math> more stars, another separator, and so on. The total number of stars in this representation is ''k'' and the number of bars is ''n'' - 1 (since no separator is needed at the very end). Thus, a string of ''k'' + ''n'' - 1 symbols (stars and bars) corresponds to a solution if there are ''k'' stars in the string. Any solution can be represented by choosing ''k'' out of {{nobreak|''k'' + ''n'' - 1}} positions to place stars and filling the remaining positions with bars. For example, the solution <math>x_1 = 3, x_2 = 2, x_3 = 0, x_4 = 5</math> of the equation <math> x_1 + x_2 + x_3 + x_4 = 10</math> can be represented by
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| <center><math>\bigstar \bigstar \bigstar | \bigstar \bigstar | | \bigstar \bigstar \bigstar \bigstar \bigstar</math>.<ref>{{harvnb|Benjamin|Quinn|2003|loc=pp. 71 –72}}</ref></center>
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| The number of such strings is the number of ways to place 10 stars in 13 positions, <math>\binom{13}{10} = \binom{13}{3} = 286,</math> which is the number of 10-multisubsets of a set with 4 elements.
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| As with binomial coefficients, there are several relationships between these multichoose expressions. For example, for <math> n \ge 1, k \ge 0</math>,
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| :<math>\left(\!\!\!\binom{n}{k}\!\!\!\right)=\left(\!\!\!\binom{k+1}{n-1}\!\!\!\right).</math>
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| This identity follows from interchanging the stars and bars in the above representation.<ref>{{harvnb|Benjamin|Quinn|2003|loc=p. 72 (identity 145)}}</ref>
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| <!--(the case where both ''r'' and ''k'' are zero is special; the correct value 1 (for the empty 0-multicombination) is given by left hand side <math>\tbinom{-1}0</math>, but not by the right hand side <math>\tbinom{-1}{-1}</math>). This follows from a clever representation of such combinations with just two symbols (see [[Stars and bars (combinatorics)]]). -->
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| === Example of counting multisubsets ===
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| For example, if you have four types of donuts (''n'' = 4) on a menu to choose from and you want three donuts (''k'' = 3), the number of ways to choose the donuts with repetition can be calculated as
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| :<math>\left(\!\!\!\binom{4}{3}\!\!\!\right) = \binom{4+3-1}3 = \binom{6}{3} = \frac{6\times5\times4}{3\times2\times1} = 20.</math>
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| This result can be verified by listing all the 3-multisubsets of the set ''S'' = {1,2,3,4}. This is displayed in the following table.<ref>{{harvnb|Benjamin|Quinn|2003|loc=p. 71}}</ref> The second column shows the nonnegative integer solutions <math>[x_1,x_2,x_3,x_4]</math> of the equation <math>x_1 + x_2 + x_3 + x_4 = 3</math> and the last column gives the stars and bars representation of the solutions.<ref>{{harvnb|Mazur|2010|loc=p. 10}} where the stars and bars are written as binary numbers, with stars = 0 and bars = 1.</ref>
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| <center>
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| {| class="wikitable"
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| |-
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| ! No.!! 3-Multiset!! Eq. Solution!! Stars and Bars
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| | 1 || {1,1,1} || [3,0,0,0] || <math>\bigstar \bigstar \bigstar |||</math>
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| |-
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| | 2 || {1,1,2} || [2,1,0,0] || <math>\bigstar \bigstar | \bigstar ||</math>
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| |-
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| | 3 || {1,1,3} || [2,0,1,0] || <math>\bigstar \bigstar ||\bigstar|</math>
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| |-
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| | 4 || {1,1,4} || [2,0,0,1] || <math>\bigstar \bigstar |||\bigstar</math>
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| |-
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| | 5 || {1,2,2} || [1,2,0,0] || <math>\bigstar |\bigstar \bigstar ||</math>
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| |-
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| | 6 || {1,2,3} || [1,1,1,0] || <math>\bigstar |\bigstar |\bigstar|</math>
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| |-
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| | 7 || {1,2,4} || [1,1,0,1] || <math>\bigstar |\bigstar ||\bigstar</math>
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| |-
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| | 8 || {1,3,3} || [1,0,2,0] || <math>\bigstar || \bigstar \bigstar |</math>
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| |-
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| | 9 || {1,3,4} || [1,0,1,1] || <math>\bigstar ||\bigstar|\bigstar</math>
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| |-
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| | 10 || {1,4,4} || [1,0,0,2] || <math>\bigstar |||\bigstar \bigstar</math>
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| |-
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| | 11 || {2,2,2} || [0,3,0,0] || <math>|\bigstar \bigstar \bigstar ||</math>
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| |-
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| | 12 || {2,2,3} || [0,2,1,0] || <math>|\bigstar \bigstar | \bigstar|</math>
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| |-
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| | 13 || {2,2,4} || [0,2,0,1] || <math>|\bigstar \bigstar ||\bigstar</math>
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| |-
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| | 14 || {2,3,3} || [0,1,2,0] || <math>|\bigstar |\bigstar \bigstar |</math>
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| |-
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| | 15 || {2,3,4} || [0,1,1,1] || <math>|\bigstar | \bigstar | \bigstar</math>
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| |-
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| | 16 || {2,4,4} || [0,1,0,2] || <math>|\bigstar ||\bigstar \bigstar</math>
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| |-
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| | 17 || {3,3,3} || [0,0,3,0] || <math>||\bigstar \bigstar \bigstar |</math>
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| |-
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| | 18 || {3,3,4} || [0,0,2,1] ||<math>||\bigstar \bigstar |\bigstar</math>
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| |-
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| | 19 || {3,4,4} || [0,0,1,2] || <math>||\bigstar |\bigstar \bigstar</math>
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| |-
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| | 20 || {4,4,4} || [0,0,0,3] || <math>|||\bigstar \bigstar \bigstar</math>
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| |}
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| </center>
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| <!--The analogy with the ''k''-combination case can be stressed by writing the numerator as a rising power
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| :<math>\binom{n + k - 1}{k} = \frac{n(n+1)\cdots(n+k-1)}{k!}.</math>
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| There is an easy way to understand the above result. Label the elements of ''S'' with numbers 0, 1, ..., {{nowrap|''n'' − 1}}, and choose a ''k''-combination from the set of numbers { 1, 2, ..., {{nowrap|''n'' + ''k'' − 1}} } (so that there are {{nowrap|''n'' − 1}} ''unchosen'' numbers). Now change this ''k''-combination into a ''k''-multicombination of ''S'' by replacing every (chosen) number ''x'' in the ''k''-combination by the element of ''S'' labeled by the ''number of unchosen numbers'' less than ''x''. This is always a number in the range of the labels, and it is easy to see that every ''k''-multicombination of ''S'' is obtained for one choice of a ''k''-combination.
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| A concrete example may be helpful. Suppose there are 4 types of fruits (apple, orange, pear, banana) at a grocery store, and you want to buy 12 pieces of fruit. So ''n'' = 4 and ''k'' = 12. Use label 0 for apples, 1 for oranges, 2 for pears, and 3 for bananas. A selection of 12 fruits can be translated into a selection of 12 distinct numbers in the range 1,...,15 by selecting as many consecutive numbers starting from 1 as there are apples in the selection, then skip a number, continue choosing as many consecutive numbers as there are oranges selected, again skip a number, then again for pears, skip one again, and finally choose the remaining numbers (as many as there are bananas selected). For instance for 2 apples, 7 oranges, 0 pears and 3 bananas, the numbers chosen will be 1, 2, 4, 5, 6, 7, 8, 9, 10, 13, 14, 15. To recover the fruits, the numbers 1, 2 (not preceded by any unchosen numbers) are replaced by apples, the numbers 4, 5, ..., 10 (preceded by one unchosen number: 3) by oranges, and the numbers 13, 14, 15 (preceded by three unchosen numbers: 3, 11, and 12) by bananas; there are no chosen numbers preceded by exactly 2 unchosen numbers, and therefore no pears in the selection.
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| The total number of possible selections is
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| <math>\binom{4+12-1}{12} = \left(\!\!\!\binom{4}{12}\!\!\!\right) = \binom{15}{12} = \left(\!\!\!\binom{13}{3}\!\!\!\right) = \binom{15}{3} = \frac{13\times14\times15}{1\times2\times3} = 455. </math>
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| -->
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| == Number of ''k''-combinations for all ''k'' ==
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| {{See also|Binomial coefficient#Sum of coefficients row}}
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| The number of ''k''-combinations for all ''k'' is the number of subsets of a set of ''n'' elements. There are several ways to see that this number is 2<sup>''n''</sup>. In terms of combinations, <math>\sum_{0\leq{k}\leq{n}}\binom nk = 2^n</math>, which is the sum of the ''n''th row (counting from 0) of the [[Binomial coefficient#Sum of coefficients row|binomial coefficients]] in [[Pascal's triangle]]. These combinations (subsets) are enumerated by the 1 digits of the set of [[base 2]] numbers counting from 0 to 2<sup>''n''</sup> - 1, where each digit position is an item from the set of ''n''.
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| Given 3 cards numbered 1 to 3, there are 8 distinct combinations ([[subsets]]), including the [[empty set]]:
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| :<math>| \{ \{\} ; \{1\} ; \{2\} ; \{3\} ; \{1, 2\} ; \{1, 3\} ; \{2, 3\} ; \{1, 2, 3\} \}| = 2^3 = 8</math>
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| Representing these subsets (in the same order) as base 2 numbers:
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| ::0 - 000
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| ::1 - 001
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| ::2 - 010
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| ::4 - 100
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| ::3 - 011
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| ::5 - 101
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| ::6 - 110
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| ::7 - 111
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| == Probability: sampling a random combination ==
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| There are various [[algorithms]] to pick out a random combination from a given set or list. [[Rejection sampling]] is extremely slow for large sample sizes. One way to select a ''k''-combination efficiently from a population of size ''n'' is to iterate across each element of the population, and at each step pick that element with a dynamically changing probability of <math>\frac{k-\mathrm{\#\,samples\ chosen}}{n-\mathrm{\#\,samples\ visited}}</math>. (see [[Reservoir_sampling|reservoir sampling]]).
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| ==See also==
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| {{colbegin|colwidth=30em}}
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| * [[Binomial coefficient]]
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| * [[Combinatorial number system]]
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| * [[Combinatorics]]
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| * [[Kneser graph]]
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| * [[List of permutation topics]]
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| * [[Multiset]]
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| * [[Pascal's triangle]]
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| * [[Permutation]]
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| * [[Probability]]
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| * [[Subset]]
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| {{colend}}
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| ==Notes==
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| {{Reflist}}
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| ==References==
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| * {{citation|first1=Arthur T.|last1=Benjamin|first2=Jennifer J.|last2=Quinn|title=Proofs that Really Count: The Art of Combinatorial Proof|year=2003|series=The Dolciani Mathematical Expositions 27|publisher=The Mathematical Association of America|isbn=978-0-88385-333-7}}
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| * {{citation|first=Richard A.|last=Brualdi|year=2010|title=Introductory Combinatorics|edition=5th|publisher=Pearson Prentice Hall|isbn=978-0-13-602040-0}}
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| * Erwin Kreyszig, ''Advanced Engineering Mathematics'', John Wiley & Sons, INC, 1999.
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| * {{citation|first=David R.|last= Mazur|title=Combinatorics: A Guided Tour|year= 2010|publisher=Mathematical Association of America|isbn=978-0-88385-762-5}}
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| * {{citation|first= Herbert John|last=Ryser|year=1963|title=Combinatorial Mathematics|publisher=Mathematical Association of America|series=The Carus Mathematical Monographs 14}}
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| ==External links==
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| * [http://community.topcoder.com/tc?module=Static&d1=tutorials&d2=combinatorics Topcoder tutorial on combinatorics ]
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| * [http://compprog.wordpress.com/2007/10/17/generating-combinations-1/ C code to generate all combinations of n elements chosen as k]
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| * [http://mathforum.org/library/drmath/sets/high_perms_combs.html Many Common types of permutation and combination math problems, with detailed solutions]
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| * [http://www.murderousmaths.co.uk/books/unknownform.htm The Unknown Formula] For combinations when choices can be repeated and order does NOT matter
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| * [http://dl.dropbox.com/u/7951257/easymath/Combinations%20with%20Repetitions.pdf Combinations with repetitions (by: Akshatha AG and Smitha B)]
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| [[Category:Combinatorics]]
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