n.

組み合わせ、併用、配合、合併、化合、コンビネーション

関: ally、combinatorial、combine、combined application、complicate、compound、concomitant、concurrent、conjunction、matching、merge、merger、mixture、pair

WordNet

the act of combining things to form a new whole (同)combining, compounding
the act of arranging elements into specified groups without regard to order
a collection of things that have been combined; an assemblage of separate parts or qualities
a coordinated sequence of chess moves
a group of people (often temporary) having a common purpose; "they were a winning combination"
a sequence of numbers or letters that opens a combination lock; "he forgot the combination to the safe"
an alliance of people or corporations or countries for a special purpose (formerly to achieve some antisocial end but now for general political or economic purposes)
form a pair or pairs; "The two old friends paired off" (同)pair off, partner off, couple
a set of two similar things considered as a unit (同)brace
a poker hand with 2 cards of the same value
two people considered as a unit
arrange in pairs; "Pair these numbers" (同)geminate
occur in pairs (同)geminate
create by mixing or combining
(chemistry) a substance formed by chemical union of two or more elements or ingredients in definite proportion by weight (同)chemical_compound
put or add together; "combine resources" (同)combine
an enclosure of residences and other building (especially in the Orient)
a whole formed by a union of two or more elements or parts
combine so as to form a whole; mix; "compound the ingredients" (同)combine
calculate principal and interest
composed of more than one part; "compound leaves are composed of several lobes; "compound flower heads"
consisting of two or more substances or ingredients or elements or parts; "soap is a compound substance"; "housetop is a compound word"; "a blackberry is a compound fruit"
join for a common purpose or in a common action; "These forces combined with others"
harvester that heads and threshes and cleans grain while moving across the field
add together from different sources; "combine resources"
the grammatical relation between linguistic units (words or phrases or clauses) that are connected by a conjunction
(astronomy) apparent meeting or passing of two or more celestial bodies in the same degree of the zodiac (同)alignment
an uninflected function word that serves to conjoin words or phrases or clauses or sentences (同)conjunctive, connective, continuative
make more complicated; "There was a new development that complicated the matter" (同)perplex
make more complex, intricate, or richer; "refine a design or pattern" (同)refine, rarify, elaborate
become an ally or associate, as by a treaty or marriage; "He allied himself with the Communists"
an associate who provides cooperation or assistance; "hes a good ally in fight" (同)friend
a friendly nation

PrepTutorEJDIC

〈U〉(…の)『結合』,組み合わせ;〈C〉(…の)組み合わされたもの,結合体《+『of』+『名』》 / 〈U〉(…との)『連合』,団結,共同,提携;〈C〉連合体,共同体,組合《+『with』+『名』》 / 〈C〉=combination lock / 《複数形で》(また《話》coms)コンビネーション(上下続きの肌着)
(二つで一つに用いられる物の)『組』,『対』《+『of』+『名』》 / (対になる部分からなる物の)『1個』《+『of』+『名』》 / 《集合的に》《複数扱い》夫婦,婚約中の男女;(動物の)つがい《+『of』+『名』》 / 《集合的に》《複数扱い》(同種または何かの関係のある物の)二人(二つ,2頭など)の組《+『of』+『名』》 / …‘を'二人(2個)ずつ組にする《+『名』+『off』(『up』),+『off』(『up』)+『名』》,(…と)…‘を'組にする《+『名』+『with』+『名』》 / (…と)…‘を'結婚させる《+『名』+[『up』]『with』+『名』》 / 二人(2個)ずつ組になる《+『off』(『up』)》
(…と)同時に起こる(発生する)《+『with』+『名』》 / (線などが)同一点に集合する / (意見などが)一致した,同一の
(…になるように)〈要素・成分など〉'を'混ぜ合わせる《+『名』+『into』+『名』》;(…から)〈薬など〉'を'調合する,混ぜ合わせて作る《+『名』+『from』(『of』)+『名』》 / 《しばしば受動態で》〈めんどうな事・損害など〉'を'もっとひどくする / 〈利子〉'を'複利で計算する / (…と)話し合いをつける,妥協する《+『with』+『名』》 / 複合の,合成の / 『合成物』,混合物,調合物;化合物 / 複合語,合成語(classroom, heart-to-heartなどをいう)
(異民族の家屋・商館などがある)囲み地区,構内
…'を'『結合する』,合同する / …'を'兼ね備える;(…と)…'を'兼ね合わさせる《+『名』+『with』+『名』》 / …'を'化合させる;(…と)…'を'化合させる《+『名』+『with』+『名』》 / 『結合する』,連合する,合同する / (…と)化合する《+『with』+『名』》 / 連合,合同;(特に)企業合同,政治連合 / (また『combine harvester』)(刈り取りと脱穀が同時にできる)複式収穫機,コンバイン
〈C〉『接続詞』(and,but,for,sinceなど) / 《文》〈U〉〈C〉結合,連結;(事件などの)同時発生 / 〈U〉合(ごう)(二つ以上の天体が同じ黄経上にあること)
…'を'複雑にする
(…と)…'を'『同盟』(連合)『させる』《+『名』+『with』(『to』)+『名』》 / 同盟国 / 味方

Wikipedia preview

出典(authority):フリー百科事典『ウィキペディア（Wikipedia）』「2014/05/23 19:10:29」(JST)

wiki en

[Wiki en表示]

In mathematics, a combination is a way of selecting members from a grouping, such that (unlike permutations) the order of selection 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. More formally, a k-combination of a 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

which can be written using factorials as whenever , and which is zero when . The set of all k-combinations of a set S is sometimes denoted by .

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,^[1] k-multiset,^[2] or k-combination with repetition are often used.^[3] 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.

With large sets, it becomes necessary to use more sophisticated mathematics to find the number of combinations. For example, a 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.

1 Number of k-combinations
- 1.1 Example of counting combinations
- 1.2 Enumerating k-combinations
2 Number of combinations with repetition
- 2.1 Example of counting multisubsets
3 Number of k-combinations for all k
4 Probability: sampling a random combination
5 See also
6 Notes
7 References
8 External links

Number of k-combinations

3-element subsets of a 5-element set

The 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 , , or even (the latter form was standard in French, Russian, Chinese^[4] and Polish texts^{[citation needed]}). The same number however occurs in many other mathematical contexts, where it is denoted by (often read as "n choose k"); notably it occurs as a coefficient in the binomial formula, hence its name binomial coefficient. One can define for all natural numbers k at once by the relation

from which it is clear that and for k > n. To see that these coefficients count k-combinations from S, one can first consider a collection of n distinct variables X_s labeled by the elements s of S, and expand the product over all elements of S:

it has 2ⁿ distinct terms corresponding to all the subsets of S, each subset giving the product of the corresponding variables X_s. Now setting all of the X_s equal to the unlabeled variable X, so that the product becomes (1 + X)ⁿ, the term for each k-combination from S becomes X^k, so that the coefficient of that power in the result equals the number of such k-combinations.

Binomial coefficients can be computed explicitly in various ways. To get all of them for the expansions up to (1 + X)ⁿ, one can use (in addition to the basic cases already given) the recursion relation

which follows from (1 + X)ⁿ = (1 + X)^{n − 1}(1 + X); this leads to the construction of Pascal's triangle.

For determining an individual binomial coefficient, it is more practical to use the formula

The numerator gives the number of 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.

When k exceeds n/2, the above formula contains factors common to the numerator and the denominator, and canceling them out gives the relation

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 of such a combination, which is an (n − k)-combination.

Finally there is a formula which exhibits this symmetry directly, and has the merit of being easy to remember:

where n! denotes the factorial of n. It is obtained from the previous formula by multiplying denominator and numerator by (n − k)!, so it is certainly inferior as a method of computation to that formula.

The last formula can be understood directly, by considering the n! 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.

From the above formulas follow relations between adjacent numbers in Pascal's triangle in all three directions:

,

.

Together with the basic cases , 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 n − k.

Example of counting combinations

As a concrete example, one can compute the number of five-card hands possible from a standard fifty-two card deck as:^[5]

{52 \choose 5} = \frac{52\times51\times50\times49\times48}{5\times4\times3\times2\times1} = \frac{311{,}875{,}200}{120} = 2{,}598{,}960.

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:

\begin{alignat}{2}{52 \choose 5} &= \frac{52!}{5!47!} \\ &= \frac{52\times51\times50\times49\times48\times\cancel{47!}}{5\times4\times3\times2\times\cancel{1}\times\cancel{47!}} \\ &= \frac{52\times51\times50\times49\times48}{5\times4\times3\times2} \\ &= \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}} \\ &= {26\times17\times10\times49\times12} \\&= 2{,}598{,}960.\end{alignat}

Another alternative computation, equivalent to the first, is based on writing

which gives

When evaluated in the following order, $52 \div 1 \times 51 \div 2 \times 50 \div 3 \times 49 \div 4 \times 48 \div 5$ , 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.

Using the symmetric formula in terms of factorials without performing simplifications gives a rather extensive calculation:

\begin{align} {52 \choose 5} &= \frac{n!}{k!(n-k)!} = \frac{52!}{5!(52-5)!} = \frac{52!}{5!47!} \\ &= \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} \\ &= 2{,}598{,}960. \end{align}

Enumerating k-combinations

One can enumerate all k-combinations of a given set $S$ of n elements in some fixed order, which establishes a bijection from an interval of integers with the set of those k-combinations. Assuming $S$ is itself ordered, for instance $S = {1,2, ..., n$ }, 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 $S$ 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.^[6]^[7]

There are many ways to enumerate k combinations. One way is to visit all the binary numbers less than . 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}.^[8]

Number of combinations with repetition

Bijection between 3-subsets of a 7-set (left)
and 3-multisets with elements from a 5-set (right)

So this illustrates that .

A k-combination with repetitions, or k-multicombination, or 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 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:^[9]

If S has n elements, the number of such k-multisubsets is denoted by,

a notation that is analogous to the binomial coefficient which counts k-subsets. This expression, n multichoose k,^[10] is also given by a binomial coefficient:

This relationship can be easily seen using a representation known as stars and bars. A solution of the above Diophantine equation can be represented by stars, a separator (a bar), then 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 k + n - 1 positions to place stars and filling the remaining positions with bars. For example, the solution of the equation can be represented by

.^[11]

The number of such strings is the number of ways to place 10 stars in 13 positions, which is the number of 10-multisubsets of a set with 4 elements.

As with binomial coefficients, there are several relationships between these multichoose expressions. For example, for ,

This identity follows from interchanging the stars and bars in the above representation.^[12]

Example of counting multisubsets

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

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.^[13] The second column shows the nonnegative integer solutions of the equation and the last column gives the stars and bars representation of the solutions.^[14]

No.	3-Multiset	Eq. Solution
1	{1,1,1}	[3,0,0,0]
2	{1,1,2}	[2,1,0,0]
3	{1,1,3}	[2,0,1,0]
4	{1,1,4}	[2,0,0,1]
5	{1,2,2}	[1,2,0,0]
6	{1,2,3}	[1,1,1,0]
7	{1,2,4}	[1,1,0,1]
8	{1,3,3}	[1,0,2,0]
9	{1,3,4}	[1,0,1,1]
10	{1,4,4}	[1,0,0,2]
11	{2,2,2}	[0,3,0,0]
12	{2,2,3}	[0,2,1,0]
13	{2,2,4}	[0,2,0,1]
14	{2,3,3}	[0,1,2,0]
15	{2,3,4}	[0,1,1,1]
16	{2,4,4}	[0,1,0,2]
17	{3,3,3}	[0,0,3,0]
18	{3,3,4}	[0,0,2,1]
19	{3,4,4}	[0,0,1,2]
20	{4,4,4}	[0,0,0,3]

Number of k-combinations for all k

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 2ⁿ. In terms of combinations, , which is the sum of the nth row (counting from 0) of the 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 2ⁿ - 1, where each digit position is an item from the set of n.

Given 3 cards numbered 1 to 3, there are 8 distinct combinations (subsets), including the empty set:

Representing these subsets (in the same order) as base 2 numbers:

0 - 000

1 - 001

2 - 010

4 - 100

3 - 011

5 - 101

6 - 110

7 - 111

Probability: sampling a random combination

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 . (see reservoir sampling).

Notes

^ Ryser 1963, p. 7 also referred to as an unordered selection.
^ Mazur 2010, p. 10
^ When the term combination is used to refer to either situation (as in (Brualdi 2010)) care must be taken to clarify whether sets or multisets are being discussed.
^ High School Textbook for full-time student (Required) Mathematics Book II B (in Chinese) (2nd ed.). China: People's Education Press. June 2006. pp. 107–116. ISBN 978-7-107-19616-4.
^ Mazur 2010, p. 21
^ http://www.site.uottawa.ca/~lucia/courses/5165-09/GenCombObj.pdf
^ http://www.sagemath.org/doc/reference/sage/combinat/subset.html
^ http://rosettacode.org/wiki/Combinations
^ Brualdi 2010, p. 52
^ Benjamin & Quinn 2003, p. 70
^ Benjamin & Quinn 2003, pp. 71 –72
^ Benjamin & Quinn 2003, p. 72 (identity 145)
^ Benjamin & Quinn 2003, p. 71
^ Mazur 2010, p. 10 where the stars and bars are written as binary numbers, with stars = 0 and bars = 1.

References

Benjamin, Arthur T.; Quinn, Jennifer J. (2003), Proofs that Really Count: The Art of Combinatorial Proof, The Dolciani Mathematical Expositions 27, The Mathematical Association of America, ISBN 978-0-88385-333-7
Brualdi, Richard A. (2010), Introductory Combinatorics (5th ed.), Pearson Prentice Hall, ISBN 978-0-13-602040-0
Erwin Kreyszig, Advanced Engineering Mathematics, John Wiley & Sons, INC, 1999.
Mazur, David R. (2010), Combinatorics: A Guided Tour, Mathematical Association of America, ISBN 978-0-88385-762-5
Ryser, Herbert John (1963), Combinatorial Mathematics, The Carus Mathematical Monographs 14, Mathematical Association of America

External links

Topcoder tutorial on combinatorics
C code to generate all combinations of n elements chosen as k
Many Common types of permutation and combination math problems, with detailed solutions
The Unknown Formula For combinations when choices can be repeated and order does NOT matter
Combinations with repetitions (by: Akshatha AG and Smitha B)

UpToDate Contents

全文を閲覧するには購読必要です。 To read the full text you will need to subscribe.

1. 糖尿病患者における高血圧治療 treatment of hypertension in patients with diabetes mellitus
2. 成人における肺高血圧症の治療 treatment of pulmonary hypertension in adults
3. 鎌状赤血球症に対するヒドロキシ尿素治療または疾患修飾療法 hydroxyurea and other disease modifying therapies in sickle cell disease
4. 最新の診療アップデート practice changing updates
5. 癌免疫療法の原則 principles of cancer immunotherapy

English Journal

Efficacy of leukotriene receptor antagonist with anti-H1 receptor antagonist plus anti-H2 receptor antagonist for treatment of refractory chronic idiopathic urticaria.

Wan KS1, Chang YS.
The Journal of dermatological treatment.J Dermatolog Treat.2014 Dec;25(6):459-61. doi: 10.3109/09546634.2013.849791. Epub 2013 Nov 5.
BACKGROUND: The efficacy of the combination of leukotriene receptor antagonist (LRA) and H1 antihistamine was similar to the synergistic regimen of H1 and H2 antihistamine for treatment of chronic idiopathic urticaria (CIU). However, the effective rates of these two regimens were only 53.3% and 63.3
PMID 24192062

Radiation-treated ready-to-eat (RTE) chicken breast Adobo for immuno-compromised patients.

Feliciano CP1, De Guzman ZM2, Tolentino LM2, Cobar ML2, Abrera GB2.
Food chemistry.Food Chem.2014 Nov 15;163:142-6. doi: 10.1016/j.foodchem.2014.04.087. Epub 2014 May 9.
Usually in hospitals low-bacterial diets are served to immuno-compromised patients (ICPs). However, low-bacterial diets still pose a high risk of microbial infections and limit the food selection of the patients. Thus, pathogen-free dishes must be made available. This study presents the development
PMID 24912709

Enzymatic process for the fractionation of baker's yeast cell wall (Saccharomyces cerevisiae).

Borchani C1, Fonteyn F2, Jamin G2, Paquot M3, Blecker C4, Thonart P2.
Food chemistry.Food Chem.2014 Nov 15;163:108-13. doi: 10.1016/j.foodchem.2014.04.086. Epub 2014 May 2.
β-Glucans, homopolymers of glucose, are widespread in many microorganisms, mushrooms and plants. They have attracted attention because of their bioactive and medicinal functions. One important source of β-glucans is the cell wall of yeasts, especially that of baker's yeast Saccharomyces cerevisiae
PMID 24912704

Incorporating measured valve properties into a numerical model of a lymphatic vessel.

Bertram CD1, Macaskill C, Moore JE Jr.
Computer methods in biomechanics and biomedical engineering.Comput Methods Biomech Biomed Engin.2014 Nov;17(14):1519-34. doi: 10.1080/10255842.2012.753066. Epub 2013 Feb 6.
An existing lumped-parameter model of multiple lymphangions (lymphatic vascular segments) in series is adapted for the incorporation of recent physiological measurements of lymphatic vascular properties. The new data show very marked nonlinearity of the passive pressure-diameter relation during dist
PMID 23387996

Japanese Journal

膝の痛み，股関節の痛み : ロコモティブシンドロームと関節

浜田大輔,Hamada Daisuke
四国医学雑誌 67(5-6), 203-204, 2012-12-25
… The causes of osteoarthritis are excessivemotion and combination such as muscle weakness and overweight.Knee osteoarthritis treatments are aimed at decreasing pain, preserving or improving jointfunction, and limiting joint deformity and disability. …
NAID 120003836502

Evolution of sex chromosomes ZW of Schistosoma mansoni inferred from chromosome paint and BAC mapping analyses.

Hirai Hirohisa,Hirai Yuriko,Loverde Philip T
Parasitology international 61(4), 684-689, 2012-12
… Combination analysis of chromosome paint analysis with Z/W probes and chromosome mapping with 93 BAC clones revealed that the W chromosome of Schistosoma mansoni has evolved by at least four inversion events and heterochromatinization. …
NAID 120004873970

A Flexible Framework for Extracting Bilingual Dictionary from Comparable Corpus Without any Language-Specific Knowledge

Xiaodong Liu,Kevin Duh,Yuji Matsumoto
情報処理学会研究報告. 自然言語処理研究会報告 2012-NL-209(15), 1-6, 2012-11-15
… Our approach is based on a novel combination of topic modeling and word alignment techniques in a pipeline style: first, our approach converts a comparable document-aligned corpus into a parallel topic-aligned corpus using topic modeling techniques, then learns translation relationships between words using word alignment models such as IBM model I. …
NAID 110009486741

A Flexible Framework for Extracting Bilingual Dictionary from Comparable Corpus Without any Language-Specific Knowledge

Xiaodong Liu,Kevin Duh,Yuji Matsumoto
情報処理学会研究報告. 自然言語処理研究会報告 2012-NL-209(15), 1-6, 2012-11-15
… Our approach is based on a novel combination of topic modeling and word alignment techniques in a pipeline style: first, our approach converts a comparable document-aligned corpus into a parallel topic-aligned corpus using topic modeling techniques, then learns translation relationships between words using word alignment models such as IBM model I. …
NAID 110009486725

Related Pictures

★リンクテーブル★

リンク元	「combine」「conjunction」「concomitant」「併用」「complicate」
拡張検索	「immunoglobulin J recombination signal sequence-binding protein」「combination antibiotic therapy」「combination of compound oxycodone and atropine」