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Wikipedia / Combination

In mathematics, a combination is a selection of items from a set that has distinct members, 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. So, two combinations are identical if and only if each combination has the same members. (The arrangement of the members in each set does not matter.) If the set has n elements, the number of k, is equal to the binomial coefficient:

which using factorial notation can be compactly expressed as

. This formula can be derived from the fact that each k-combination of a set S of n. The set of all k-combinations of a set S.

A combination is a selection of n things taken k at a time without repetition. To refer to combinations in which repetition is allowed, the terms k-combination with repetition, k-multiset, or k-selection, 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 is1/2,598,960.

Number of k-combinations

The number of k-combinations from a given set S of n. (often read as "n choose k"); notably it occurs as a coefficient in the binomial formula for all natural numbers k at once by the relation

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 ofS:

it has 2n 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 , the term for each k-combination from S becomes Xk, 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 , one can use (in addition to the basic cases already given) the recursion relation

for 0 < k < n, which follows from =; 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 -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 !, 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 (nk) 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:

, 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 .

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:

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

which gives

When evaluated in the following order, , 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

One can enumerate all k-combinations of a given set S of n elements in some fixed order, which establishes a bijection 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 at0, 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.

There are many ways to enumerate k combinations. One 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. Then, repeatedly move to the next allowed k-combination by incrementing the smallest index number for which this would not create two equal index numbers, at the same time resetting all smaller index numbers to their initial values.

Number of combinations with repetition

A k-combination with repetitions, or k-multicombination, or multisubset of size k from a set S of size n is given by a set of k not necessarily distinct elements of S, where order is not taken into account: two sequences define the same multiset if one can be obtained from the other by permuting the terms. In other words, it is a sample of 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 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 (so allowing zero) solutions of the Diophantine equation:

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. stars, a separator (a bar 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 a separation into n parts needs n-1 separators). Thus, a string of k + n - 1 (or n + k - 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 (n = 4 and k = 10) can be represented by

which is the number of 10-multisubsets of a set with 4 elements.

,

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

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. and the last column gives the stars and bars representation of the solutions.

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, 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 numerals:

  • 0 – 000

  • 1 – 001

  • 2 – 010

  • 3 – 011

  • 4 – 100

  • 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 (see Reservoir sampling and convert it into a combination using the combinatorial number system.

Number of ways to put objects into bins

A combination can also be thought of as a selection of two sets of items: those that go into the chosen bin and those that go into the unchosen bin. This can be generalized to any number of bins with the constraint that every item must go to exactly one bin. The number of ways to put objects into bins is given by the multinomial coefficient

where n is the number of items, m is the number of items that go into bin i.

One way to see why this equation holds is to first number the objects arbitrarily from 1 to n distinct numberings, but many of them are equivalent, because only the set of items in a bin matters, not their order in it. Every combined permutation of each bins' contents produces an equivalent way of putting items into bins. As a result, every equivalence class.

The binomial coefficient is the special case where k items go into the unchosen bin:

See also

Notes

References

  • Erwin Kreyszig, Advanced Engineering Mathematics, John Wiley & Sons, INC, 1999.