Английская Википедия:Binomial coefficient

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Файл:Pascal's triangle 5.svg

The binomial coefficients can be arranged to form Pascal's triangle, in which each entry is the sum of the two immediately above.

Файл:Binomial theorem visualisation.svg

Visualisation of binomial expansion up to the 4th power

In mathematics, the binomial coefficients are the positive integers that occur as coefficients in the binomial theorem. Commonly, a binomial coefficient is indexed by a pair of integers Шаблон:Math and is written <math>\tbinom{n}{k}.</math> It is the coefficient of the Шаблон:Math term in the polynomial expansion of the binomial power Шаблон:Math; this coefficient can be computed by the multiplicative formula

<math>\binom nk = \frac{n\times(n-1)\times\cdots\times(n-k+1)}{k\times(k-1)\times\cdots\times1},</math>

which using factorial notation can be compactly expressed as

<math>\binom{n}{k} = \frac{n!}{k! (n-k)!}.</math>

For example, the fourth power of Шаблон:Math is

<math>\begin{align}

(1 + x)^4 &= \tbinom{4}{0} x^0 + \tbinom{4}{1} x^1 + \tbinom{4}{2} x^2 + \tbinom{4}{3} x^3 + \tbinom{4}{4} x^4 \\ &= 1 + 4x + 6 x^2 + 4x^3 + x^4, \end{align}</math> and the binomial coefficient <math>\tbinom{4}{2} =\tfrac{4\times 3}{2\times1} = \tfrac{4!}{2!2!} = 6</math> is the coefficient of the Шаблон:Math term.

Arranging the numbers <math>\tbinom{n}{0}, \tbinom{n}{1}, \ldots, \tbinom{n}{n}</math> in successive rows for Шаблон:Math gives a triangular array called Pascal's triangle, satisfying the recurrence relation

<math>\binom{n}{k} = \binom{n-1}{k-1} + \binom{n-1}{k} .</math>

The binomial coefficients occur in many areas of mathematics, and especially in combinatorics. The symbol <math>\tbinom{n}{k}</math> is usually read as "Шаблон:Mvar choose Шаблон:Mvar" because there are <math>\tbinom{n}{k}</math> ways to choose an (unordered) subset of Шаблон:Mvar elements from a fixed set of Шаблон:Mvar elements. For example, there are <math>\tbinom{4}{2}=6</math> ways to choose Шаблон:Math elements from Шаблон:Math, namely Шаблон:Math, Шаблон:Math, Шаблон:Math, Шаблон:Math, Шаблон:Math and Шаблон:Math.

The binomial coefficients can be generalized to <math>\tbinom{z}{k}</math> for any complex number Шаблон:Mvar and integer Шаблон:Math, and many of their properties continue to hold in this more general form.

History and notation

Andreas von Ettingshausen introduced the notation <math>\tbinom nk</math> in 1826, although the numbers were known centuries earlier (see Pascal's triangle). In about 1150, the Indian mathematician Bhaskaracharya gave an exposition of binomial coefficients in his book Līlāvatī.

Alternative notations include Шаблон:Math, Шаблон:Math, Шаблон:Math, Шаблон:Math, Шаблон:Math, and Шаблон:Math, in all of which the Шаблон:Mvar stands for combinations or choices. Many calculators use variants of the Шаблон:Nowrap because they can represent it on a single-line display. In this form the binomial coefficients are easily compared to [[permutation#k-permutations of n|Шаблон:Mvar-permutations of Шаблон:Mvar]], written as Шаблон:Math, etc.

Definition and interpretations

Шаблон:Diagonal split header	0	1	2	3	4	⋯
0	1	Шаблон:SilverC	Шаблон:SilverC	Шаблон:SilverC	Шаблон:SilverC	⋯
1	1	1	Шаблон:SilverC	Шаблон:SilverC	Шаблон:SilverC	⋯
2	1	2	1	Шаблон:SilverC	Шаблон:SilverC	⋯
3	1	3	3	1	Шаблон:SilverC	⋯
4	1	4	6	4	1	⋯
⋮	⋮	⋮	⋮	⋮	⋮	⋱
The first few binomial coefficients on a left-aligned Pascal's triangle

For natural numbers (taken to include 0) n and k, the binomial coefficient <math>\tbinom nk</math> can be defined as the coefficient of the monomial X^k in the expansion of Шаблон:Math. The same coefficient also occurs (if Шаблон:Math) in the binomial formula Шаблон:NumBlk (valid for any elements x, y of a commutative ring), which explains the name "binomial coefficient".

Another occurrence of this number is in combinatorics, where it gives the number of ways, disregarding order, that Шаблон:Mvar objects can be chosen from among Шаблон:Mvar objects; more formally, the number of Шаблон:Mvar-element subsets (or Шаблон:Mvar-combinations) of an Шаблон:Mvar-element set. This number can be seen as equal to the one of the first definition, independently of any of the formulas below to compute it: if in each of the Шаблон:Mvar factors of the power Шаблон:Math one temporarily labels the term Шаблон:Mvar with an index Шаблон:Mvar (running from Шаблон:Math to Шаблон:Mvar), then each subset of Шаблон:Mvar indices gives after expansion a contribution Шаблон:Math, and the coefficient of that monomial in the result will be the number of such subsets. This shows in particular that <math>\tbinom nk</math> is a natural number for any natural numbers Шаблон:Mvar and Шаблон:Mvar. There are many other combinatorial interpretations of binomial coefficients (counting problems for which the answer is given by a binomial coefficient expression), for instance the number of words formed of Шаблон:Mvar bits (digits 0 or 1) whose sum is Шаблон:Mvar is given by <math>\tbinom nk</math>, while the number of ways to write <math>k = a_1 + a_2 + \cdots + a_n</math> where every Шаблон:Math is a nonnegative integer is given by Шаблон:Tmath. Most of these interpretations can be shown to be equivalent to counting Шаблон:Mvar-combinations.

Computing the value of binomial coefficients

Several methods exist to compute the value of <math>\tbinom{n}{k}</math> without actually expanding a binomial power or counting Шаблон:Mvar-combinations.

Recursive formula

One method uses the recursive, purely additive formula <math display="block"> \binom nk = \binom{n-1}{k-1} + \binom{n-1}k</math> for all integers <math>n,k</math> such that <math>1 \le k < n,</math> with boundary values <math display="block">\binom n0 = \binom nn = 1</math> for all integers Шаблон:Math.

The formula follows from considering the set Шаблон:Math and counting separately (a) the Шаблон:Mvar-element groupings that include a particular set element, say "Шаблон:Mvar", in every group (since "Шаблон:Mvar" is already chosen to fill one spot in every group, we need only choose Шаблон:Math from the remaining Шаблон:Math) and (b) all the k-groupings that don't include "Шаблон:Mvar"; this enumerates all the possible Шаблон:Mvar-combinations of Шаблон:Mvar elements. It also follows from tracing the contributions to X^k in Шаблон:Math. As there is zero Шаблон:Math or Шаблон:Math in Шаблон:Math, one might extend the definition beyond the above boundaries to include <math>\tbinom nk = 0</math> when either Шаблон:Math or Шаблон:Math. This recursive formula then allows the construction of Pascal's triangle, surrounded by white spaces where the zeros, or the trivial coefficients, would be.

Multiplicative formula

A more efficient method to compute individual binomial coefficients is given by the formula <math display="block">\binom nk = \frac{n^{\underline{k}}}{k!} = \frac{n(n-1)(n-2)\cdots(n-(k-1))}{k(k-1)(k-2)\cdots 1} = \prod_{i=1}^k\frac{ n+1-i}{ i},</math> where the numerator of the first fraction <math>n^{\underline{k}}</math> is expressed as a falling factorial power. This formula is easiest to understand for the combinatorial interpretation of binomial coefficients. The numerator gives the number of ways to select a sequence of Шаблон:Mvar distinct objects, retaining the order of selection, from a set of Шаблон:Mvar objects. The denominator counts the number of distinct sequences that define the same Шаблон:Mvar-combination when order is disregarded.

Due to the symmetry of the binomial coefficient with regard to Шаблон:Mvar and Шаблон:Math, calculation may be optimised by setting the upper limit of the product above to the smaller of Шаблон:Mvar and Шаблон:Math.

Factorial formula

Finally, though computationally unsuitable, there is the compact form, often used in proofs and derivations, which makes repeated use of the familiar factorial function: <math display="block"> \binom nk = \frac{n!}{k!\,(n-k)!} \quad \text{for }\ 0\leq k\leq n,</math> where Шаблон:Math denotes the factorial of Шаблон:Mvar. This formula follows from the multiplicative formula above by multiplying numerator and denominator by Шаблон:Math; as a consequence it involves many factors common to numerator and denominator. It is less practical for explicit computation (in the case that Шаблон:Mvar is small and Шаблон:Mvar is large) unless common factors are first cancelled (in particular since factorial values grow very rapidly). The formula does exhibit a symmetry that is less evident from the multiplicative formula (though it is from the definitions) Шаблон:NumBlk which leads to a more efficient multiplicative computational routine. Using the falling factorial notation, <math display="block"> \binom nk = \begin{cases}

n^{\underline{k}}/k! & \text{if }\ k \le \frac{n}{2} \\
n^{\underline{n-k}}/(n-k)! & \text{if }\ k > \frac{n}{2}

\end{cases}. </math>

Generalization and connection to the binomial series

Шаблон:Main The multiplicative formula allows the definition of binomial coefficients to be extended by replacing n by an arbitrary number α (negative, real, complex) or even an element of any commutative ring in which all positive integers are invertible: <math display="block">\binom \alpha k = \frac{\alpha^{\underline k}}{k!} = \frac{\alpha(\alpha-1)(\alpha-2)\cdots(\alpha-k+1)}{k(k-1)(k-2)\cdots 1} \quad\text{for } k\in\N \text{ and arbitrary } \alpha. </math>

With this definition one has a generalization of the binomial formula (with one of the variables set to 1), which justifies still calling the <math>\tbinom\alpha k</math> binomial coefficients: Шаблон:NumBlk

This formula is valid for all complex numbers α and X with |X| < 1. It can also be interpreted as an identity of formal power series in X, where it actually can serve as definition of arbitrary powers of power series with constant coefficient equal to 1; the point is that with this definition all identities hold that one expects for exponentiation, notably <math display="block">(1+X)^\alpha(1+X)^\beta=(1+X)^{\alpha+\beta} \quad\text{and}\quad ((1+X)^\alpha)^\beta=(1+X)^{\alpha\beta}.</math>

If α is a nonnegative integer n, then all terms with Шаблон:Math are zero, and the infinite series becomes a finite sum, thereby recovering the binomial formula. However, for other values of α, including negative integers and rational numbers, the series is really infinite.

Pascal's triangle

Файл:Pascal's triangle - 1000th row.png

1000th row of Pascal's triangle, arranged vertically, with grey-scale representations of decimal digits of the coefficients, right-aligned. The left boundary of the image corresponds roughly to the graph of the logarithm of the binomial coefficients, and illustrates that they form a log-concave sequence.

Pascal's rule is the important recurrence relation Шаблон:NumBlk which can be used to prove by mathematical induction that <math> \tbinom n k</math> is a natural number for all integer n ≥ 0 and all integer k, a fact that is not immediately obvious from formula (1). To the left and right of Pascal's triangle, the entries (shown as blanks) are all zero.

Pascal's rule also gives rise to Pascal's triangle:

0:									1
1:								1		1
2:							1		2		1
3:						1		3		3		1
4:					1		4		6		4		1
5:				1		5		10		10		5		1
6:			1		6		15		20		15		6		1
7:		1		7		21		35		35		21		7		1
8:	1		8		28		56		70		56		28		8		1

Row number Шаблон:Mvar contains the numbers <math>\tbinom{n}{k}</math> for Шаблон:Math. It is constructed by first placing 1s in the outermost positions, and then filling each inner position with the sum of the two numbers directly above. This method allows the quick calculation of binomial coefficients without the need for fractions or multiplications. For instance, by looking at row number 5 of the triangle, one can quickly read off that

<math>(x + y)^5 = \underline{1}x^5 + \underline{5}x^4y + \underline{10}x^3y^2 + \underline{10}x^2y^3 + \underline{5}xy^4 + \underline{1}y^5.</math>

Combinatorics and statistics

Binomial coefficients are of importance in combinatorics, because they provide ready formulas for certain frequent counting problems:

• There are <math>\tbinom n k</math> ways to choose k elements from a set of n elements. See Combination.
• There are <math>\tbinom {n+k-1}k</math> ways to choose k elements from a set of n elements if repetitions are allowed. See Multiset.
• There are <math> \tbinom {n+k} k</math> strings containing k ones and n zeros.
• There are <math> \tbinom {n+1} k</math> strings consisting of k ones and n zeros such that no two ones are adjacent.
• The Catalan numbers are <math>\tfrac{1}{n+1}\tbinom{2n}{n}.</math>
• The binomial distribution in statistics is <math>\tbinom n k p^k (1-p)^{n-k} .</math>

Binomial coefficients as polynomials

For any nonnegative integer k, the expression <math display="inline">\binom{t}{k}</math> can be simplified and defined as a polynomial divided by Шаблон:Math:

<math>\binom{t}{k} = \frac{t^\underline{k}}{k!} = \frac{t(t-1)(t-2)\cdots(t-k+1)}{k(k-1)(k-2)\cdots2 \cdot 1};</math>

this presents a polynomial in t with rational coefficients.

As such, it can be evaluated at any real or complex number t to define binomial coefficients with such first arguments. These "generalized binomial coefficients" appear in Newton's generalized binomial theorem.

For each k, the polynomial <math>\tbinom{t}{k}</math> can be characterized as the unique degree k polynomial Шаблон:Math satisfying Шаблон:Math and Шаблон:Math.

Its coefficients are expressible in terms of Stirling numbers of the first kind:

<math>\binom{t}{k} = \sum_{i=0}^k s(k,i)\frac{t^i}{k!}.</math>

The derivative of <math>\tbinom{t}{k}</math> can be calculated by logarithmic differentiation:

<math>\frac{\mathrm{d}}{\mathrm{d}t} \binom{t}{k} = \binom{t}{k} \sum_{i=0}^{k-1} \frac{1}{t-i}.</math>

This can cause a problem when evaluated at integers from <math>0</math> to <math>t-1</math>, but using identities below we can compute the derivative as:

<math>\frac{\mathrm{d}}{\mathrm{d}t} \binom{t}{k} = \sum_{i=0}^{k-1} \frac{(-1)^{k-i-1}}{k-i} \binom{t}{i}.</math>

Binomial coefficients as a basis for the space of polynomials

Over any field of characteristic 0 (that is, any field that contains the rational numbers), each polynomial p(t) of degree at most d is uniquely expressible as a linear combination <math display="inline">\sum_{k=0}^d a_k \binom{t}{k}</math> of binomial coefficients. The coefficient a_k is the kth difference of the sequence p(0), p(1), ..., p(k). Explicitly, Шаблон:NumBlk

Integer-valued polynomials

Шаблон:Main Each polynomial <math>\tbinom{t}{k}</math> is integer-valued: it has an integer value at all integer inputs <math>t</math>. (One way to prove this is by induction on k, using Pascal's identity.) Therefore, any integer linear combination of binomial coefficient polynomials is integer-valued too. Conversely, (Шаблон:EquationNote) shows that any integer-valued polynomial is an integer linear combination of these binomial coefficient polynomials. More generally, for any subring R of a characteristic 0 field K, a polynomial in K[t] takes values in R at all integers if and only if it is an R-linear combination of binomial coefficient polynomials.

Example

The integer-valued polynomial Шаблон:Math can be rewritten as

<math>9\binom{t}{2} + 6 \binom{t}{1} + 0\binom{t}{0}.</math>

Identities involving binomial coefficients

The factorial formula facilitates relating nearby binomial coefficients. For instance, if k is a positive integer and n is arbitrary, then Шаблон:NumBlk and, with a little more work,

<math>\binom {n-1}{k} - \binom{n-1}{k-1} = \frac{n-2k}{n} \binom{n}{k}.</math>

We can also get

<math>\binom {n-1}{k} = \frac{n-k}{n} \binom {n}{k}.</math>

Moreover, the following may be useful:

<math>\binom{n}{h}\binom{n-h}{k}=\binom{n}{k}\binom{n-k}{h}=\binom{n}{h+k}\binom{h+k}{h}.</math>

For constant n, we have the following recurrence:

<math> \binom{n}{k} = \frac{n-k+1}{k} \binom{n}{k-1}.</math>

To sum up, we have

<math>\binom {n}{k} = \binom n{n-k} = \frac{n-k+1}{k} \binom {n}{k-1} = \frac{n}{n-k} \binom {n-1}{k} = \frac{n}{k} \binom {n-1}{k-1} = \frac{n}{n-2k} \Bigg(\binom {n-1}{k} - \binom{n-1}{k-1}\Bigg) = \binom{n-1}k + \binom{n-1}{k-1}.</math>

Sums of the binomial coefficients

The formula Шаблон:NumBlk says that the elements in the Шаблон:Mvarth row of Pascal's triangle always add up to 2 raised to the Шаблон:Mvarth power. This is obtained from the binomial theorem (Шаблон:EquationNote) by setting Шаблон:Math and Шаблон:Math. The formula also has a natural combinatorial interpretation: the left side sums the number of subsets of {1, ..., n} of sizes k = 0, 1, ..., n, giving the total number of subsets. (That is, the left side counts the power set of {1, ..., n}.) However, these subsets can also be generated by successively choosing or excluding each element 1, ..., n; the n independent binary choices (bit-strings) allow a total of <math>2^n</math> choices. The left and right sides are two ways to count the same collection of subsets, so they are equal.

The formulas Шаблон:NumBlk and

<math>\sum_{k=0}^n k^2 \binom n k = (n + n^2)2^{n-2}</math>

follow from the binomial theorem after differentiating with respect to Шаблон:Mvar (twice for the latter) and then substituting Шаблон:Math.

The Chu–Vandermonde identity, which holds for any complex values m and n and any non-negative integer k, is Шаблон:NumBlk and can be found by examination of the coefficient of <math>x^k</math> in the expansion of Шаблон:Math using equation (Шаблон:EquationNote). When Шаблон:Math, equation (Шаблон:EquationNote) reduces to equation (Шаблон:EquationNote). In the special case Шаблон:Math, using (Шаблон:EquationNote), the expansion (Шаблон:EquationNote) becomes (as seen in Pascal's triangle at right) Шаблон:Image frame Шаблон:NumBlk where the term on the right side is a central binomial coefficient.

Another form of the Chu–Vandermonde identity, which applies for any integers j, k, and n satisfying Шаблон:Math, is Шаблон:NumBlk The proof is similar, but uses the binomial series expansion (Шаблон:EquationNote) with negative integer exponents. When Шаблон:Math, equation (Шаблон:EquationNote) gives the hockey-stick identity

<math>\sum_{m=k}^n \binom m k = \binom {n+1}{k+1}</math>

and its relative

<math>\sum_{r=0}^m \binom {n+r} r = \binom {n+m+1}{m}.</math>

Let F(n) denote the n-th Fibonacci number. Then

<math> \sum_{k=0}^{\lfloor n/2\rfloor} \binom {n-k} k = F(n+1).</math>

This can be proved by induction using (Шаблон:EquationNote) or by Zeckendorf's representation. A combinatorial proof is given below.

Multisections of sums

For integers s and t such that <math>0\leq t < s,</math> series multisection gives the following identity for the sum of binomial coefficients:

<math>\binom{n}{t}+\binom{n}{t+s}+\binom{n}{t+2s}+\ldots=\frac{1}{s}\sum_{j=0}^{s-1}\left(2\cos\frac{\pi j}{s}\right)^n\cos\frac{\pi(n-2t)j}{s}.</math>

For small Шаблон:Mvar, these series have particularly nice forms; for example,

<math> \binom{n}{0} + \binom{n}{3}+\binom{n}{6}+\cdots = \frac{1}{3}\left(2^n +2 \cos\frac{n\pi}{3}\right) </math>

<math> \binom{n}{1} + \binom{n}{4}+\binom{n}{7}+\cdots = \frac{1}{3}\left(2^n +2 \cos\frac{(n-2)\pi}{3}\right) </math>

<math> \binom{n}{2} + \binom{n}{5}+\binom{n}{8}+\cdots = \frac{1}{3}\left(2^n +2 \cos\frac{(n-4)\pi}{3}\right) </math>

<math> \binom{n}{0} + \binom{n}{4}+\binom{n}{8}+\cdots = \frac{1}{2}\left(2^{n-1} +2^{\frac{n}{2}} \cos\frac{n\pi}{4}\right) </math>

<math> \binom{n}{1} + \binom{n}{5}+\binom{n}{9}+\cdots = \frac{1}{2}\left(2^{n-1} +2^{\frac{n}{2}} \sin\frac{n\pi}{4}\right) </math>

<math> \binom{n}{2} + \binom{n}{6}+\binom{n}{10}+\cdots = \frac{1}{2}\left(2^{n-1} -2^{\frac{n}{2}} \cos\frac{n\pi}{4}\right) </math>

<math> \binom{n}{3} + \binom{n}{7}+\binom{n}{11}+\cdots = \frac{1}{2}\left(2^{n-1} -2^{\frac{n}{2}} \sin\frac{n\pi}{4}\right) </math>

Partial sums

Although there is no closed formula for partial sums

<math> \sum_{j=0}^k \binom n j</math>

of binomial coefficients, one can again use (Шаблон:EquationNote) and induction to show that for Шаблон:Math,

<math> \sum_{j=0}^k (-1)^j\binom{n}{j} = (-1)^k\binom {n-1}{k},</math>

with special case

<math>\sum_{j=0}^n (-1)^j\binom n j = 0</math>

for Шаблон:Math. This latter result is also a special case of the result from the theory of finite differences that for any polynomial P(x) of degree less than n,

<math> \sum_{j=0}^n (-1)^j\binom n j P(j) = 0.</math>

Differentiating (Шаблон:EquationNote) k times and setting x = −1 yields this for <math>P(x)=x(x-1)\cdots(x-k+1)</math>, when 0 ≤ k < n, and the general case follows by taking linear combinations of these.

When P(x) is of degree less than or equal to n, Шаблон:NumBlk where <math>a_n</math> is the coefficient of degree n in P(x).

More generally for (Шаблон:EquationNote),

<math> \sum_{j=0}^n (-1)^j\binom n j P(m+(n-j)d) = d^n n! a_n</math>

where m and d are complex numbers. This follows immediately applying (Шаблон:EquationNote) to the polynomial Шаблон:Tmath instead of Шаблон:Tmath, and observing that Шаблон:Tmath still has degree less than or equal to n, and that its coefficient of degree n is dⁿa_n.

The series <math display="inline">\frac{k-1}{k} \sum_{j=0}^\infty \frac 1 {\binom {j+x} k}= \frac 1 {\binom{x-1}{k-1}}</math> is convergent for k ≥ 2. This formula is used in the analysis of the German tank problem. It follows from <math display="inline">\frac{k-1}k\sum_{j=0}^{M}\frac 1 {\binom{j+x} k}=\frac 1{\binom{x-1}{k-1}}-\frac 1{\binom{M+x}{k-1}}</math> which is proved by induction on M.

Identities with combinatorial proofs

Many identities involving binomial coefficients can be proved by combinatorial means. For example, for nonnegative integers <math>{n} \geq {q}</math>, the identity

<math>\sum_{k=q}^n \binom{n}{k} \binom{k}{q} = 2^{n-q}\binom{n}{q}</math>

(which reduces to (Шаблон:EquationNote) when q = 1) can be given a double counting proof, as follows. The left side counts the number of ways of selecting a subset of [n] = {1, 2, ..., n} with at least q elements, and marking q elements among those selected. The right side counts the same thing, because there are <math>\tbinom n q</math> ways of choosing a set of q elements to mark, and <math>2^{n-q}</math> to choose which of the remaining elements of [n] also belong to the subset.

In Pascal's identity

<math>{n \choose k} = {n-1 \choose k-1} + {n-1 \choose k},</math>

both sides count the number of k-element subsets of [n]: the two terms on the right side group them into those that contain element n and those that do not.

The identity (Шаблон:EquationNote) also has a combinatorial proof. The identity reads

<math>\sum_{k=0}^n \binom{n}{k}^2 = \binom{2n}{n}.</math>

Suppose you have <math>2n</math> empty squares arranged in a row and you want to mark (select) n of them. There are <math>\tbinom {2n}n</math> ways to do this. On the other hand, you may select your n squares by selecting k squares from among the first n and <math>n-k</math> squares from the remaining n squares; any k from 0 to n will work. This gives

<math>\sum_{k=0}^n\binom n k\binom n{n-k} = \binom {2n} n.</math>

Now apply (Шаблон:EquationNote) to get the result.

If one denotes by Шаблон:Math the sequence of Fibonacci numbers, indexed so that Шаблон:Math, then the identity <math display="block">\sum_{k=0}^{\left\lfloor\frac{n}{2}\right\rfloor} \binom {n-k} k = F(n)</math> has the following combinatorial proof. One may show by induction that Шаблон:Math counts the number of ways that a Шаблон:Math strip of squares may be covered by Шаблон:Math and Шаблон:Math tiles. On the other hand, if such a tiling uses exactly Шаблон:Mvar of the Шаблон:Math tiles, then it uses Шаблон:Math of the Шаблон:Math tiles, and so uses Шаблон:Math tiles total. There are <math>\tbinom{n-k}{k}</math> ways to order these tiles, and so summing this coefficient over all possible values of Шаблон:Mvar gives the identity.

Sum of coefficients row

The number of k-combinations for all k, <math display="inline">\sum_{0\leq{k}\leq{n}}\binom nk = 2^n</math>, is the sum of the nth row (counting from 0) of the binomial coefficients. These combinations are enumerated by the 1 digits of the set of base 2 numbers counting from 0 to <math>2^n - 1</math>, where each digit position is an item from the set of n.

Dixon's identity

Dixon's identity is

<math>\sum_{k=-a}^{a}(-1)^{k}{2a\choose k+a}^3 =\frac{(3a)!}{(a!)^3}</math>

or, more generally,

<math>\sum_{k=-a}^a(-1)^k{a+b\choose a+k} {b+c\choose b+k}{c+a\choose c+k} = \frac{(a+b+c)!}{a!\,b!\,c!}\,,</math>

where a, b, and c are non-negative integers.

Continuous identities

Certain trigonometric integrals have values expressible in terms of binomial coefficients: For any <math>m, n \in \N,</math>

<math>\int_{-\pi}^{\pi} \cos((2m-n)x)\cos^n(x)\ dx = \frac{\pi}{2^{n-1}} \binom{n}{m}</math>

<math>

\int_{-\pi}^{\pi} \sin((2m-n)x)\sin^n(x)\ dx = \begin{cases} (-1)^{m+(n+1)/2} \frac{\pi}{2^{n-1}} \binom{n}{m}, & n \text{ odd} \\ 0, & \text{otherwise} \end{cases}</math>

<math>

\int_{-\pi}^{\pi} \cos((2m-n)x)\sin^n(x)\ dx = \begin{cases} (-1)^{m+(n/2)} \frac{\pi}{2^{n-1}} \binom{n}{m}, & n \text{ even} \\ 0, & \text{otherwise} \end{cases}</math>

These can be proved by using Euler's formula to convert trigonometric functions to complex exponentials, expanding using the binomial theorem, and integrating term by term.

Congruences

If n is prime, then <math display="block">\binom {n-1}k \equiv (-1)^k \mod n</math> for every k with <math>0\leq k \leq n-1.</math> More generally, this remains true if n is any number and k is such that all the numbers between 1 and k are coprime to n.

Indeed, we have

<math>

\binom {n-1} k = {(n-1)(n-2)\cdots(n-k)\over 1\cdot 2\cdots k} = \prod_{i=1}^{k}{n-i\over i}\equiv \prod_{i=1}^{k}{-i\over i} = (-1)^k \mod n. </math>

Generating functions

Ordinary generating functions

For a fixed Шаблон:Mvar, the ordinary generating function of the sequence <math>\tbinom n0,\tbinom n1,\tbinom n2,\ldots</math> is

<math>\sum_{k=0}^\infty {n\choose k} x^k = (1+x)^n.</math>

For a fixed Шаблон:Mvar, the ordinary generating function of the sequence <math>\tbinom 0k,\tbinom 1k, \tbinom 2k,\ldots,</math> is

<math>\sum_{n=0}^\infty {n\choose k} y^n = \frac{y^k}{(1-y)^{k+1}}.</math>

The bivariate generating function of the binomial coefficients is

<math>\sum_{n=0}^\infty \sum_{k=0}^n {n\choose k} x^k y^n = \frac{1}{1-y-xy}.</math>

A symmetric bivariate generating function of the binomial coefficients is

<math>\sum_{n=0}^\infty \sum_{k=0}^\infty {n+k\choose k} x^k y^n = \frac{1}{1-x-y}.</math>

which is the same as the previous generating function after the substitution <math>x\to xy</math>.

Exponential generating function

A symmetric exponential bivariate generating function of the binomial coefficients is:

<math>\sum_{n=0}^\infty \sum_{k=0}^\infty {n+k\choose k} \frac{x^k y^n}{(n+k)!} = e^{x+y}.</math>

Divisibility properties

Шаблон:Main In 1852, Kummer proved that if m and n are nonnegative integers and p is a prime number, then the largest power of p dividing <math>\tbinom{m+n}{m}</math> equals p^c, where c is the number of carries when m and n are added in base p. Equivalently, the exponent of a prime p in <math>\tbinom n k</math> equals the number of nonnegative integers j such that the fractional part of k/p^j is greater than the fractional part of n/p^j. It can be deduced from this that <math>\tbinom n k</math> is divisible by n/gcd(n,k). In particular therefore it follows that p divides <math>\tbinom{p^r}{s}</math> for all positive integers r and s such that Шаблон:Math. However this is not true of higher powers of p: for example 9 does not divide <math>\tbinom{9}{6}</math>.

A somewhat surprising result by David Singmaster (1974) is that any integer divides almost all binomial coefficients. More precisely, fix an integer d and let f(N) denote the number of binomial coefficients <math>\tbinom n k</math> with n < N such that d divides <math>\tbinom n k</math>. Then

<math> \lim_{N\to\infty} \frac{f(N)}{N(N+1)/2} = 1. </math>

Since the number of binomial coefficients <math>\tbinom n k</math> with n < N is N(N + 1) / 2, this implies that the density of binomial coefficients divisible by d goes to 1.

Binomial coefficients have divisibility properties related to least common multiples of consecutive integers. For example:

<math>\binom{n+k}k</math> divides <math>\frac{\operatorname{lcm}(n,n+1,\ldots,n+k)}n</math>.

<math>\binom{n+k}k</math> is a multiple of <math>\frac{\operatorname{lcm}(n,n+1,\ldots,n+k)}{n\cdot \operatorname{lcm}(\binom{k}0,\binom{k}1,\ldots,\binom{k}k)}</math>.

Another fact: An integer Шаблон:Math is prime if and only if all the intermediate binomial coefficients

<math> \binom n 1, \binom n 2, \ldots, \binom n{n-1} </math>

are divisible by n.

Proof: When p is prime, p divides

<math> \binom p k = \frac{p \cdot (p-1) \cdots (p-k+1)}{k \cdot (k-1) \cdots 1} </math> for all Шаблон:Math

because <math>\tbinom p k</math> is a natural number and p divides the numerator but not the denominator. When n is composite, let p be the smallest prime factor of n and let Шаблон:Math. Then Шаблон:Math and

<math> \binom n p = \frac{n(n-1)(n-2)\cdots(n-p+1)}{p!}=\frac{k(n-1)(n-2)\cdots(n-p+1)}{(p-1)!}\not\equiv 0 \pmod{n}</math>

otherwise the numerator Шаблон:Math has to be divisible by Шаблон:Math, this can only be the case when Шаблон:Math is divisible by p. But n is divisible by p, so p does not divide Шаблон:Math and because p is prime, we know that p does not divide Шаблон:Math and so the numerator cannot be divisible by n.

Bounds and asymptotic formulas

The following bounds for <math>\tbinom n k</math> hold for all values of n and k such that Шаблон:Math: <math display="block">\frac{n^k}{k^k} \le {n \choose k} \le \frac{n^k}{k!} < \left(\frac{n\cdot e}{k}\right)^k.</math> The first inequality follows from the fact that <math display="block"> {n \choose k} = \frac{n}{k} \cdot \frac{n-1}{k-1} \cdots \frac{n-(k-1)}{1} </math> and each of these <math>k </math> terms in this product is <math display="inline"> \geq \frac{n}{k} </math>. A similar argument can be made to show the second inequality. The final strict inequality is equivalent to <math display="inline">e^k > k^k / k!</math>, that is clear since the RHS is a term of the exponential series <math display="inline"> e^k = \sum_{j=0}^\infty k^j/j! </math>.

From the divisibility properties we can infer that <math display="block">\frac{\operatorname{lcm}(n-k, \ldots, n)}{(n-k) \cdot \operatorname{lcm}\left(\binom{k}{0}, \ldots, \binom{k}{k}\right)}\leq\binom{n}{k} \leq \frac{\operatorname{lcm}(n-k, \ldots, n)}{n - k},</math> where both equalities can be achieved.

The following bounds are useful in information theory:Шаблон:Rp <math display="block"> \frac{1}{n+1} 2^{nH(k/n)} \leq {n \choose k} \leq 2^{nH(k/n)} </math> where <math>H(p) = -p\log_2(p) -(1-p)\log_2(1-p)</math> is the binary entropy function. It can be further tightened to <math display="block"> \sqrt{\frac{n}{8k(n-k)}} 2^{nH(k/n)} \leq {n \choose k} \leq \sqrt{\frac{n}{2\pi k(n-k)}} 2^{nH(k/n)}</math> for all <math>1 \leq k \leq n-1</math>.Шаблон:Rp

Both n and k large

Stirling's approximation yields the following approximation, valid when <math>n-k,k</math> both tend to infinity: <math display="block">{n \choose k} \sim \sqrt{n\over 2\pi k (n-k)} \cdot {n^n \over k^k (n-k)^{n-k}} </math> Because the inequality forms of Stirling's formula also bound the factorials, slight variants on the above asymptotic approximation give exact bounds. In particular, when <math>n</math> is sufficiently large, one has <math display="block"> {2n \choose n} \sim \frac{2^{2n}}{\sqrt{n\pi }}</math> and <math>\sqrt{n}{2n \choose n} \ge 2^{2n-1}</math> and, more generally, for Шаблон:Math and Шаблон:Math,Шаблон:Why <math display="block">\sqrt{n}{mn \choose n} \ge \frac{m^{m(n-1)+1}}{(m-1)^{(m-1)(n-1)}}.</math>

If n is large and k is linear in n, various precise asymptotic estimates exist for the binomial coefficient <math display="inline"> \binom{n}{k}</math>. For example, if <math>| n/2 - k | = o(n^{2/3})</math> then <math display="block"> \binom{n}{k} \sim \binom{n}{\frac{n}{2}} e^{-d^2/(2n)} \sim \frac{2^n}{\sqrt{\frac{1}{2}n \pi }} e^{-d^2/(2n)}</math> where d = n − 2k.

Шаблон:Mvar much larger than Шаблон:Mvar

If Шаблон:Mvar is large and Шаблон:Mvar is Шаблон:Math (that is, if Шаблон:Math), then <math display = block> \binom{n}{k} \sim \left(\frac{n e}{k} \right)^k \cdot (2\pi k)^{-1/2} \cdot \exp\left(- \frac{k^2}{2n}(1 + o(1))\right)</math> where again Шаблон:Mvar is the little o notation.

Sums of binomial coefficients

A simple and rough upper bound for the sum of binomial coefficients can be obtained using the binomial theorem: <math display="block">\sum_{i=0}^k {n \choose i} \leq \sum_{i=0}^k n^i\cdot 1^{k-i} \leq (1+n)^k</math> More precise bounds are given by <math display="block">\frac{1}{\sqrt{8n\varepsilon(1-\varepsilon)}} \cdot 2^{H(\varepsilon) \cdot n} \leq \sum_{i=0}^{k} \binom{n}{i} \leq 2^{H(\varepsilon) \cdot n},</math> valid for all integers <math>n > k \geq 1</math> with <math>\varepsilon \doteq k/n \leq 1/2</math>.

Generalized binomial coefficients

The infinite product formula for the gamma function also gives an expression for binomial coefficients <math display="block">(-1)^k {z \choose k}= {-z+k-1 \choose k} = \frac{1}{\Gamma(-z)} \frac{1}{(k+1)^{z+1}} \prod_{j=k+1} \frac{\left(1+\frac{1}{j}\right)^{-z-1}}{1-\frac{z+1}{j}}</math> which yields the asymptotic formulas <math display="block">{z \choose k} \approx \frac{(-1)^k}{\Gamma(-z) k^{z+1}} \qquad \text{and} \qquad {z+k \choose k} = \frac{k^z}{\Gamma(z+1)}\left( 1+\frac{z(z+1)}{2k}+\mathcal{O}\left(k^{-2}\right)\right)</math> as <math>k \to \infty</math>.

This asymptotic behaviour is contained in the approximation <math display="block">{z+k \choose k}\approx \frac{e^{z(H_k-\gamma)}}{\Gamma(z+1)}</math> as well. (Here <math>H_k</math> is the k-th harmonic number and <math>\gamma</math> is the Euler–Mascheroni constant.)

Further, the asymptotic formula <math display="block">\fracШаблон:Z+k\choose jШаблон:K\choose j\to \left(1-\frac{j}{k}\right)^{-z}\quad\text{and}\quad \fracШаблон:J\choose j-kШаблон:J-z\choose j-k\to \left(\frac{j}{k}\right)^z</math> hold true, whenever <math>k\to\infty</math> and <math>j/k \to x</math> for some complex number <math>x</math>.

Generalizations

Generalization to multinomials

Шаблон:Main Binomial coefficients can be generalized to multinomial coefficients defined to be the number:

<math>{n\choose k_1,k_2,\ldots,k_r} =\frac{n!}{k_1!k_2!\cdots k_r!}</math>

where

<math>\sum_{i=1}^rk_i=n.</math>

While the binomial coefficients represent the coefficients of Шаблон:Math, the multinomial coefficients represent the coefficients of the polynomial

<math>(x_1 + x_2 + \cdots + x_r)^n.</math>

The case r = 2 gives binomial coefficients:

<math>{n\choose k_1,k_2}={n\choose k_1, n-k_1}={n\choose k_1}= {n\choose k_2}.</math>

The combinatorial interpretation of multinomial coefficients is distribution of n distinguishable elements over r (distinguishable) containers, each containing exactly k_i elements, where i is the index of the container.

Multinomial coefficients have many properties similar to those of binomial coefficients, for example the recurrence relation:

<math>{n\choose k_1,k_2,\ldots,k_r} ={n-1\choose k_1-1,k_2,\ldots,k_r}+{n-1\choose k_1,k_2-1,\ldots,k_r}+\ldots+{n-1\choose k_1,k_2,\ldots,k_r-1}</math>

and symmetry:

<math>{n\choose k_1,k_2,\ldots,k_r} ={n\choose k_{\sigma_1},k_{\sigma_2},\ldots,k_{\sigma_r}}</math>

where <math>(\sigma_i)</math> is a permutation of (1, 2, ..., r).

Taylor series

Using Stirling numbers of the first kind the series expansion around any arbitrarily chosen point <math>z_0</math> is

<math>\begin{align} {z \choose k} = \frac{1}{k!}\sum_{i=0}^k z^i s_{k,i}&=\sum_{i=0}^k (z- z_0)^i \sum_{j=i}^k {z_0 \choose j-i} \frac{s_{k+i-j,i}}{(k+i-j)!} \\ &=\sum_{i=0}^k (z-z_0)^i \sum_{j=i}^k z_0^{j-i} {j \choose i} \frac{s_{k,j}}{k!}.\end{align}</math>

Binomial coefficient with Шаблон:Math

The definition of the binomial coefficients can be extended to the case where <math>n</math> is real and <math>k</math> is integer.

In particular, the following identity holds for any non-negative integer <math>k</math>:

<math>{{1/2}\choose{k}}={{2k}\choose{k}}\frac{(-1)^{k+1}}{2^{2k}(2k-1)}.</math>

This shows up when expanding <math>\sqrt{1+x}</math> into a power series using the Newton binomial series :

<math>\sqrt{1+x}=\sum_{k\geq 0}{\binom{1/2}{k}}x^k.</math>

Products of binomial coefficients

One can express the product of two binomial coefficients as a linear combination of binomial coefficients:

<math>{z \choose m} {z\choose n} = \sum_{k=0}^{\min(m,n)} {m + n - k \choose k, m - k, n - k} {z \choose m + n - k},</math>

where the connection coefficients are multinomial coefficients. In terms of labelled combinatorial objects, the connection coefficients represent the number of ways to assign Шаблон:Math labels to a pair of labelled combinatorial objects—of weight m and n respectively—that have had their first k labels identified, or glued together to get a new labelled combinatorial object of weight Шаблон:Math. (That is, to separate the labels into three portions to apply to the glued part, the unglued part of the first object, and the unglued part of the second object.) In this regard, binomial coefficients are to exponential generating series what falling factorials are to ordinary generating series.

The product of all binomial coefficients in the nth row of the Pascal triangle is given by the formula:

<math>\prod_{k=0}^{n}\binom{n}{k}=\prod_{k=1}^{n}k^{2k-n-1}.</math>

Partial fraction decomposition

The partial fraction decomposition of the reciprocal is given by

<math>\frac{1}Шаблон:Z \choose n= \sum_{i=0}^{n-1} (-1)^{n-1-i} {n \choose i} \frac{n-i}{z-i},

\qquad \frac{1}Шаблон:Z+n \choose n= \sum_{i=1}^n (-1)^{i-1} {n \choose i} \frac{i}{z+i}.</math>

Newton's binomial series

Шаблон:Main Newton's binomial series, named after Sir Isaac Newton, is a generalization of the binomial theorem to infinite series:

<math> (1+z)^{\alpha} = \sum_{n=0}^{\infty}{\alpha\choose n}z^n = 1+{\alpha\choose1}z+{\alpha\choose 2}z^2+\cdots.</math>

The identity can be obtained by showing that both sides satisfy the differential equation Шаблон:Math.

The radius of convergence of this series is 1. An alternative expression is

<math>\frac{1}{(1-z)^{\alpha+1}} = \sum_{n=0}^{\infty}{n+\alpha \choose n}z^n</math>

where the identity

<math>{n \choose k} = (-1)^k {k-n-1 \choose k}</math>

is applied.

Multiset (rising) binomial coefficient

Шаблон:Main Binomial coefficients count subsets of prescribed size from a given set. A related combinatorial problem is to count multisets of prescribed size with elements drawn from a given set, that is, to count the number of ways to select a certain number of elements from a given set with the possibility of selecting the same element repeatedly. The resulting numbers are called multiset coefficients; the number of ways to "multichoose" (i.e., choose with replacement) k items from an n element set is denoted <math display="inline">\left(\!\!\binom n k\!\!\right)</math>.

To avoid ambiguity and confusion with n's main denotation in this article,
let Шаблон:Math and Шаблон:Math.

Multiset coefficients may be expressed in terms of binomial coefficients by the rule <math display="block">\binom{f}{k}=\left(\!\!\binom{r}{k}\!\!\right)=\binom{r+k-1}{k}.</math> One possible alternative characterization of this identity is as follows: We may define the falling factorial as <math display="block">(f)_{k}=f^{\underline k}=(f-k+1)\cdots(f-3)\cdot(f-2)\cdot(f-1)\cdot f,</math> and the corresponding rising factorial as <math display="block">r^{(k)}=\,r^{\overline k}=\,r\cdot(r+1)\cdot(r+2)\cdot(r+3)\cdots(r+k-1);</math> so, for example, <math display="block">17\cdot18\cdot19\cdot20\cdot21=(21)_{5}=21^{\underline 5}=17^{\overline 5}=17^{(5)}.</math> Then the binomial coefficients may be written as <math display="block">\binom{f}{k} = \frac{(f)_{k}}{k!} =\frac{(f-k+1)\cdots(f-2)\cdot(f-1)\cdot f}{1\cdot2\cdot3\cdot4\cdot5\cdots k} ,</math> while the corresponding multiset coefficient is defined by replacing the falling with the rising factorial: <math display="block">\left(\!\!\binom{r}{k}\!\!\right)=\frac{r^{(k)}}{k!}=\frac{r\cdot(r+1)\cdot(r+2)\cdots(r+k-1)}{1\cdot2\cdot3\cdot4\cdot5\cdots k}.</math>

Generalization to negative integers n

Шаблон:Pascal triangle extended.svg For any n,

<math>\begin{align}\binom{-n}{k} &= \frac{-n\cdot-(n+1)\dots-(n+k-2)\cdot-(n+k-1)}{k!}\\

&=(-1)^k\;\frac{n\cdot(n+1)\cdot(n+2)\cdots (n + k - 1)}{k!}\\ &=(-1)^k\binom{n + k - 1}{k}\\ &=(-1)^k\left(\!\!\binom{n}{k}\!\!\right)\;.\end{align}</math> In particular, binomial coefficients evaluated at negative integers n are given by signed multiset coefficients. In the special case <math>n = -1</math>, this reduces to <math>(-1)^k=\binom{-1}{k}=\left(\!\!\binom{-k}{k}\!\!\right) .</math>

For example, if n = −4 and k = 7, then r = 4 and f = 10:

<math>\begin{align}\binom{-4}{7} &= \frac

{-10\cdot-9\cdot-8\cdot-7\cdot-6\cdot-5\cdot-4} {1\cdot2\cdot3\cdot4\cdot5\cdot6\cdot7}\\ &=(-1)^7\;\frac{4\cdot5\cdot6\cdot7\cdot8\cdot9\cdot10} {1\cdot2\cdot3\cdot4\cdot5\cdot6\cdot7}\\ &=\left(\!\!\binom{-7}{7}\!\!\right)\left(\!\!\binom{4}{7}\!\!\right)=\binom{-1}{7}\binom{10}{7}.\end{align}</math>

Two real or complex valued arguments

The binomial coefficient is generalized to two real or complex valued arguments using the gamma function or beta function via

<math>{x \choose y}= \frac{\Gamma(x+1)}{\Gamma(y+1) \Gamma(x-y+1)}= \frac{1}{(x+1) \Beta(y+1, x-y+1)}.</math>

This definition inherits these following additional properties from <math>\Gamma</math>:

<math>{x \choose y}= \frac{\sin (y \pi)}{\sin(x \pi)} {-y-1 \choose -x-1}= \frac{\sin((x-y) \pi)}{\sin (x \pi)} {y-x-1 \choose y};</math>

moreover,

<math>{x \choose y} \cdot {y \choose x}= \frac{\sin((x-y) \pi)}{(x-y) \pi}.</math>

The resulting function has been little-studied, apparently first being graphed in Шаблон:Harv. Notably, many binomial identities fail: <math display="inline">\binom{n }{ m} = \binom{n }{ n-m}</math> but <math display="inline">\binom{-n}{m} \neq \binom{-n}{-n-m}</math> for n positive (so <math>-n</math> negative). The behavior is quite complex, and markedly different in various octants (that is, with respect to the x and y axes and the line <math>y=x</math>), with the behavior for negative x having singularities at negative integer values and a checkerboard of positive and negative regions:

• in the octant <math>0 \leq y \leq x</math> it is a smoothly interpolated form of the usual binomial, with a ridge ("Pascal's ridge").
• in the octant <math>0 \leq x \leq y</math> and in the quadrant <math>x \geq 0, y \leq 0</math> the function is close to zero.
• in the quadrant <math>x \leq 0, y \geq 0</math> the function is alternatingly very large positive and negative on the parallelograms with vertices <math display="block">(-n,m+1), (-n,m), (-n-1,m-1), (-n-1,m)</math>
• in the octant <math>0 > x > y</math> the behavior is again alternatingly very large positive and negative, but on a square grid.
• in the octant <math>-1 > y > x + 1</math> it is close to zero, except for near the singularities.

Generalization to q-series

The binomial coefficient has a q-analog generalization known as the Gaussian binomial coefficient.

Generalization to infinite cardinals

The definition of the binomial coefficient can be generalized to infinite cardinals by defining:

<math>{\alpha \choose \beta} = \left| \left\{ B \subseteq A : \left|B\right| = \beta \right\} \right|</math>

where Шаблон:Mvar is some set with cardinality <math>\alpha</math>. One can show that the generalized binomial coefficient is well-defined, in the sense that no matter what set we choose to represent the cardinal number <math>\alpha</math>, <math display="inline">{\alpha \choose \beta}</math> will remain the same. For finite cardinals, this definition coincides with the standard definition of the binomial coefficient.

Assuming the Axiom of Choice, one can show that <math display="inline">{\alpha \choose \alpha} = 2^{\alpha}</math> for any infinite cardinal <math>\alpha</math>.

See also

• Binomial transform
• Delannoy number
• Eulerian number
• Hypergeometric function
• List of factorial and binomial topics
• Macaulay representation of an integer
• Motzkin number
• Multiplicities of entries in Pascal's triangle
• Narayana number
• Star of David theorem
• Sun's curious identity
• Table of Newtonian series
• Trinomial expansion

References

• Шаблон:Cite book
• Шаблон:Cite book
• Шаблон:Cite book
• Шаблон:Cite book
• Шаблон:Cite journal
• Шаблон:Cite journal
• Шаблон:Cite book
• Шаблон:Cite book
• Шаблон:Citation
• Шаблон:Cite book
• Шаблон:Cite book
• Шаблон:Cite journal
• Шаблон:Cite book
• Шаблон:Citation

External links

• Шаблон:Springer
• Шаблон:Cite journal