Combination
This article is about the mathematics of selecting part of a collection.
For other uses, see Combination (disambiguation).
"COMBIN" and "nCr" redirect here.
For other uses, see Combin (disambiguation) and NCR (disambiguation).
In mathematics, a combination is a selection of items from a collection, such that the order of selection does not matter (unlike permutations).
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
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, k-multiset, or k-combination with repetition are often used.
If, in the above example, it were 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.
Although the set of three fruits was small enough to write a complete list of combinations, this becomes impractical as the size of the set increases.
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.
Number of k-combinations
Main article: Binomial coefficient
from which it is clear that
and further,
To see that these coefficients count k-combinations from S, one can first consider a collection of n distinct variables Xs labeled by the elements s of S, and expand the product over all elements of S:
it has 2 distinct terms corresponding to all the subsets of S, each subset giving the product of the corresponding variables Xs.
Now setting all of the Xs equal to the unlabeled variable X, so that the product becomes (1 + X), the term for each k-combination from S becomes X, 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), one can use (in addition to the basic cases already given) the recursion relation
for 0 < k < n, which follows from (1 + X) = (1 + X)(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
for 0 ≤ k ≤ n. 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 computationally less efficient than 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:
Example of counting combinations
As a specific example, one can compute the number of five-card hands possible from a standard fifty-two card deck as:
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:
Another alternative computation, equivalent to the first, is based on writing
which gives
When evaluated in the following order, 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.
Using the symmetric formula in terms of factorials without performing simplifications gives a rather extensive calculation:
Enumerating k-combinations
There are many ways to enumerate k combinations.
One way is to visit all the binary numbers less than 2.
Choose those numbers having k nonzero bits, although this is very inefficient even for small n (e.g. n = 20 would require visiting about one million numbers while the maximum number of allowed k combinations is about 186 thousand for k = 10).
The positions of these 1 bits in such a number is a specific k-combination of the set { 1, …, n }.
Another simple, faster way is to track k index numbers of the elements selected, starting with {0 .. k−1} (zero-based) or {1 .. k} (one-based) as the first allowed k-combination and then repeatedly moving to the next allowed k-combination by incrementing the last index number if it is lower than n-1 (zero-based) or n (one-based) or the last index number x that is less than the index number following it minus one if such an index exists and resetting the index numbers after x to {x+1, x+2, …}.
Number of combinations with repetition
See also: Multiset coefficient
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, can also be given in terms of binomial coefficients:
This relationship can be easily proved using a representation known as stars and bars.
This identity follows from interchanging the stars and bars in the above representation.
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
Number of k-combinations for all k
See also: Binomial coefficient § Sum of coefficients row
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 numerals:
- 0 – 000
- 1 – 001
- 2 – 010
- 3 – 011
- 4 – 100
- 5 – 101
- 6 – 110
- 7 – 111
Probability: sampling a random combination
See also
Credits to the contents of this page go to the authors of the corresponding Wikipedia page: en.wikipedia.org/wiki/Combination.