Combination

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 C^n_k, {}_nC_k, {}^nC_k, C_{n,k} or even C_n^k (the last form is standard in French, Romanian, Russian, and Chinese texts). The same number however occurs in many other mathematical contexts, where it is denoted by \tbinom nk (often read as "n choose k"); notably it occurs as a coefficient in the binomial formula, hence its name binomial coefficient. One can define \tbinom nk for all natural numbers k at once by the relation (1 + X)^n = \sum_{k\geq0}\binom{n}{k} X^k, from which it is clear that \binom{n}{0} = \binom{n}{n} = 1, and further \binom{n}{k} = 0 for k>n. 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: \prod_{s\in S}(1+X_s); 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 \binom{n}{k} = \binom{n - 1}{k - 1} + \binom{n - 1}{k}, for 0 \binom nk = \frac{n(n-1)(n-2)\cdots(n-k+1)}{k!}. 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 \binom nk = \binom n{n-k}, 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: \binom nk = \frac{n!}{k!(n-k)!}, 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 (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: \binom nk = \begin{cases} \displaystyle \binom n{k-1} \frac {n-k+1}k &\quad \text{if } k > 0 \\ \displaystyle \binom {n-1}k \frac n{n-k} &\quad \text{if } k 0 \end{cases}. Together with the basic cases \tbinom n0=1=\tbinom nn, 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: \binom{52}{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} \binom{52}{5} &= \frac{52!}{5!47!} \\[5pt] &= \frac{52\times51\times50\times49\times48\times\cancel{47!}}{5\times4\times3\times2\times\cancel{1}\times\cancel{47!}} \\[5pt] &= \frac{52\times51\times50\times49\times48}{5\times4\times3\times2} \\[5pt] &= \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}} \\[5pt] &= {26\times17\times10\times49\times12} \\[5pt] &= 2{,}598{,}960. \end{alignat} Another alternative computation, equivalent to the first, is based on writing \binom{n}{k} = \frac { ( n - 0 ) }1 \times \frac { ( n - 1 ) }2 \times \frac { ( n - 2 ) }3 \times \cdots \times \frac { ( n - (k - 1) ) }k, which gives \binom{52}{5} = \frac{52}1 \times \frac{51}2 \times \frac{50}3 \times \frac{49}4 \times \frac{48}5 = 2{,}598{,}960. 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: \begin{align} \binom{52}{5} &= \frac{n!}{k!(n-k)!} = \frac{52!}{5!(52-5)!} = \frac{52!}{5!47!} \\[6pt] &= \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} \\[6pt] &= 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 \tbinom nk 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. 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 of objects, then we can let x_i 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: x_1 + x_2 + \ldots + x_n = k. If S has n elements, the number of such k-multisubsets is denoted by \left(\!\!\binom{n}{k}\!\!\right), 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: \left(\!\!\binom{n}{k}\!\!\right)=\binom{n+k-1}{k}. This relationship can be easily proved using a representation known as stars and bars. A solution of the above Diophantine equation can be represented by x_1 stars, a separator (a bar), then x_2 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 positions to place stars and filling the remaining positions with bars. For example, the solution x_1 = 3, x_2 = 2, x_3 = 0, x_4 = 5 of the equation x_1 + x_2 + x_3 + x_4 = 10 (n = 4 and k = 10) can be represented by \bigstar \bigstar \bigstar | \bigstar \bigstar | | \bigstar \bigstar \bigstar \bigstar \bigstar. The number of such strings is the number of ways to place 10 stars in 13 positions, \binom{13}{10} = \binom{13}{3} = 286, which is the number of 10-multisubsets of a set with 4 elements. between 3-subsets of a 7-set (left) and 3-multisets with elements from a 5-set (right).This illustrates that \binom{7}{3} = \left(\!\! \binom{5}{3}\!\!\right). As with binomial coefficients, there are several relationships between these multichoose expressions. For example, for n \ge 1, k \ge 0, \left(\!\!\binom{n}{k}\!\!\right)=\left(\!\!\binom{k+1}{n-1}\!\!\right). This identity follows from interchanging the stars and bars in the above representation. \tbinom{-1}0, but not by the right hand side \tbinom{-1}{-1}). This follows from a clever representation of such combinations with just two symbols (see Stars and bars (combinatorics)). --> 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 \left(\!\!\binom{4}{3}\!\!\right) = \binom{4+3-1}3 = \binom{6}{3} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20. 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. The second column lists the donuts you actually chose, the third column shows the nonnegative integer solutions [x_1,x_2,x_3,x_4] of the equation x_1 + x_2 + x_3 + x_4 = 3 and the last column gives the stars and bars representation of the solutions. \binom{n + k - 1}{k} = \frac{n(n+1)\cdots(n+k-1)}{k!}. There is an easy way to understand the above result. Label the elements of S with numbers 0, 1, ..., , and choose a k-combination from the set of numbers { 1, 2, ..., } (so that there are 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. 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. The total number of possible selections is \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. --> == 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, \sum_{0\leq{k}\leq{n}}\binom n k = 2^n, 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: | \{ \{\} ; \{1\} ; \{2\} ; \{1, 2\} ; \{3\} ; \{1, 3\} ; \{2, 3\} ; \{1, 2, 3\} \}| = 2^3 = 8 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 is to iterate across each element of the population, and at each step pick that element with a dynamically changing probability of \frac{k-\#\text{samples chosen}}{n- \#\text{samples visited}} (see Reservoir sampling). Another is to pick a random non-negative integer less than \textstyle\binom nk 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 \binom{n}{k_1, k_2, \ldots, k_m} = \frac{n!}{k_1!\, k_2! \cdots k_m!}, where n is the number of items, m is the number of bins, and k_i 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 and put the objects with numbers 1, 2, \ldots, k_1 into the first bin in order, the objects with numbers k_1+1, k_1+2, \ldots, k_1+k_2 into the second bin in order, and so on. There are 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 consists of k_1!\, k_2! \cdots k_m! distinct numberings, and the number of equivalence classes is \textstyle\frac{n!}{k_1!\, k_2! \cdots k_m!}. The binomial coefficient is the special case where k items go into the chosen bin and the remaining n-k items go into the unchosen bin: \binom nk = \binom{n}{k, n-k} = \frac{n!}{k!(n-k)!}. ==See also==