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[[Image:Pascal's triangle 5.svg|right|thumb|200px|The binomial coefficients can be arranged to form [[Pascal's triangle]].]]
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In [[mathematics]], '''binomial coefficients''' are a family of positive [[integer]]s that occur as [[coefficient]]s in the [[binomial theorem]]. They are indexed by two nonnegative integers; the binomial coefficient indexed by ''n'' and ''k'' is usually written <math>\tbinom nk</math>. It is the [[coefficient]] of the ''x''<sup>&nbsp;''k''</sup> term in the [[polynomial expansion]] of the [[binomial]] [[exponentiation|power]] (1&nbsp;+&nbsp;''x'')<sup>&nbsp;''n''</sup>. Under suitable circumstances the value of the coefficient is given by the expression <math>\tfrac{n!}{k!\,(n-k)!}</math>. Arranging binomial coefficients into rows for successive values of ''n'', and in which  ''k'' ranges from 0 to ''n'', gives a triangular array called [[Pascal's triangle]].
 
This family of numbers also arises in many other areas than algebra, notably in [[combinatorics]]. For any set containing ''n'' elements, the number of distinct ''k''-element [[subset]]s of it that can be formed (the ''k''-[[combination]]s of its elements) is given by the binomial coefficient <math>\tbinom nk</math>. Therefore <math>\tbinom nk</math> is often read as "''n'' choose ''k''". The properties of binomial coefficients have led to extending the meaning of the symbol <math>\tbinom nk</math> beyond the basic case where ''n'' and ''k'' are nonnegative integers with {{nowrap|''k'' ≤ ''n''}}; such expressions are then still called binomial coefficients.
 
The notation <math>\tbinom nk</math> was introduced by [[Andreas von Ettingshausen]] in 1826,<ref>{{harvtxt|Higham|1998}}</ref> although the numbers were already known centuries before that (see [[Pascal's triangle]]). The earliest known detailed discussion of binomial coefficients is in a tenth-century commentary, by [[Halayudha]], on an ancient [[Sanskrit]] text, [[Pingala]]'s Chandaḥśāstra.  In about 1150, the Indian mathematician [[Bhaskaracharya]] gave a very clear exposition of binomial coefficients in his book [[Lilavati]].<ref>[[Lilavati]] Section 6, Chapter 4 (see {{harvtxt|Knuth|1997}}).</ref>
 
Alternative notations include ''C''(''n'', ''k''), <sub>''n''</sub>''C''<sub>''k''</sub>, <sup>''n''</sup>''C''<sub>''k''</sub>, ''C''<sup>''k''</sup><sub style="position:relative; left:-.5em;">''n''</sub>, ''C''<sup>''n''</sup><sub style="position:relative; left:-.5em;">''k''</sub>,<ref>{{harvtxt|Shilov|1977}}</ref> ''C''<sub>''n'',''k''</sub> in all of which the ''C'' stands for ''[[combination]]s'' or ''choices''.
 
== Definition and interpretations ==
For [[natural number]]s (taken to include 0) ''n'' and ''k'', the binomial coefficient <math>\tbinom nk</math> can be defined as the [[coefficient]] of the [[monomial]] ''X''<sup>''k''</sup> in the expansion of {{nowrap|(1 + ''X'')<sup>''n''</sup>}}. The same coefficient also occurs (if {{nowrap|''k'' ≤ ''n''}}) in the [[binomial formula]]
:<math>(x+y)^n=\sum_{k=0}^n\binom nk x^{n-k}y^k</math>
(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 ''k'' objects can be chosen from among ''n'' objects; more formally, the number of ''k''-element subsets (or ''k''-[[combination]]s) of an ''n''-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 ''n'' factors of the power {{nowrap|(1 + ''X'')<sup>''n''</sup>}} one temporarily labels the term ''X'' with an index ''i'' (running from 1 to ''n''), then each subset of ''k'' indices gives after expansion a contribution ''X''<sup>''k''</sup>, 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 ''n'' and ''k''. 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 ''n'' [[bit]]s (digits 0 or 1) whose sum is ''k'' 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 ''a''<sub>''i''</sub> is a nonnegative integer is given by <math>\tbinom{n+k-1}{n-1}</math>. Most of these interpretations are easily seen to be equivalent to counting ''k''-combinations.
 
== Computing the value of binomial coefficients ==
Several methods exist to compute the value of <math>\tbinom nk</math> without actually expanding a binomial power or counting ''k''-combinations.
 
=== Recursive formula ===
One method uses the [[recursion|recursive]], purely additive, formula
 
:<math>  \binom nk = \binom{n-1}{k-1} + \binom{n-1}k \quad \text{for all integers }n,k : 1\le k\le n-1,</math>
 
with initial/boundary values
 
:<math>\binom n0 = \binom nn = 1 \quad \text{for all integers } n\ge0,</math>
 
The formula follows from considering the set {1,2,3,…,''n''} and counting separately (a) the ''k''-element groupings that include a particular set element, say “''i''”, in every group (since “''i''” is already chosen to fill one spot in every group, we need only choose ''k''&nbsp;−&nbsp;1 from the remaining ''n''&nbsp;&minus;&nbsp;1) and (b) all the ''k''-groupings that don’t include “''i''”; this enumerates all the possible ''k''-combinations of ''n'' elements. It also follows from tracing the contributions to ''X''<sup>''k''</sup> in {{nowrap|(1 + ''X'')<sup>''n''−1</sup>(1 + ''X'')}}. As there is zero ''X''<sup>''n''+1</sup> or ''X''<sup>&minus;1</sup> in {{nowrap|(1 + ''X'')<sup>''n''</sup>}}, one might extend the definition beyond the above boundaries to include <math>\tbinom nk</math>&nbsp;=&nbsp;0 when either ''k''&nbsp;>&nbsp;''n'' or ''k''&nbsp;<&nbsp;0. 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>\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-(k-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 ''k'' distinct objects, retaining the order of selection, from a set of ''n'' objects. The denominator counts the number of distinct sequences that define the same ''k''-combination when order is disregarded.
 
=== 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> \binom nk = \frac{n!}{k!\,(n-k)!} \quad \text{for }\ 0\leq k\leq n,</math>
 
where ''n''! denotes the factorial of ''n''. This formula follows from the multiplicative formula above by multiplying numerator and denominator by {{nowrap|(''n'' − ''k'')!}}; as a consequence it involves many factors common to numerator and denominator. It is less practical for explicit computation 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)
 
{{NumBlk|:|<math> \binom nk = \binom n{n-k} \quad \text{for }\ 0\leq k\leq n,</math>|{{EquationRef|1}}}}
 
which leads to a more efficient multiplicative computational routine. Using the falling factorial notation,
 
:<math> \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 ===
The multiplicative formula allows the definition of binomial coefficients to be extended<ref>See {{Harv|Graham|Knuth|Patashnik|1994}}, which also defines <math>\tbinom n k = 0</math> for <math>k<0</math>. Alternative generalizations, such as to [[#Two real or complex valued arguments|two real or complex valued arguments]] using the [[Gamma function]] assign nonzero values to <math>\tbinom n k</math> for <math>k<0</math>, but this causes most binomial coefficient identities to fail, and thus is not widely used majority of definitions.  One such choice of nonzero values leads to the aesthetically pleasing "Pascal windmill" in Hilton, Holton and Pedersen, ''Mathematical reflections: in a room with many mirrors'', Springer, 1997, but causes even [[Pascal's identity]] to fail (at the origin).</ref> 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>\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:
 
{{NumBlk|:|<math> (1+X)^\alpha = \sum_{k=0}^\infty {\alpha \choose k} X^k.</math>|{{EquationRef|2}}}}
 
This formula is valid for all complex numbers ''α'' and ''X'' with |''X''|&nbsp;&lt;&nbsp;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 series with constant coefficient equal to&nbsp;1; the point is that with this definition all identities hold that one expects for [[exponentiation]], notably
 
:<math>(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 ''k''&nbsp;>&nbsp;''n'' 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 ==
[[Image:Pascal's triangle - 1000th row.png|150px|right|thumb|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 [[logarithmically concave sequence|log-concave sequence]].]]
 
{{Main|Pascal's triangle|Pascal's rule}}
 
[[Pascal's rule]] is the important [[recurrence relation]]
{{NumBlk|:|<math> {n \choose k} +  {n \choose k+1} = {n+1 \choose k+1},</math>|{{EquationRef|3}}}}
which can be used to prove by [[mathematical induction]] that <math> \tbinom n k</math> is a natural number for all ''n'' and ''k'', (equivalent to the statement that k! divides the product of k consecutive integers),  a fact that is not immediately obvious from [[#Definition_and_interpretations|formula (1)]].
 
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&nbsp;|| ||7&nbsp;|| ||21|| ||35|| ||35|| ||21|| ||7&nbsp;|| ||1&nbsp;||
|-
|8: ||1&nbsp;|| ||8&nbsp;|| ||28|| ||56|| ||70|| ||56|| ||28|| ||8&nbsp;|| ||1&nbsp;
|} <!--There is a wider cell made with &nbsp; in 1-digit columns, so triangle becomes more graphically symmetrical -->
 
Row number ''n'' contains the numbers <math> \tbinom n k</math> for ''k'' = 0,…,''n''. It is constructed by starting with ones at the outside and then always adding two adjacent numbers and writing the sum directly underneath. 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
:(''x'' + ''y'')<sup>5</sup> = '''1''' ''x''<sup>5</sup> + '''5''' ''x''<sup>4</sup>''y'' + '''10''' ''x''<sup>3</sup>''y''<sup>2</sup> + '''10''' ''x''<sup>2</sup>''y''<sup>3</sup> + '''5''' ''x'' ''y''<sup>4</sup> + '''1''' ''y''<sup>5</sup>.
The differences between elements on other diagonals are the elements in the previous diagonal, as a consequence of the recurrence relation ({{EquationNote|3}}) above.
 
== 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> [[string (computer science)|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.<ref>{{cite journal|last=Muir|first=Thomas|title=Note on Selected Combinations|journal=Proceedings of the Royal Society of Edinburgh|year=1902|url=http://books.google.com/books/reader?id=EN8vAAAAIAAJ&output=reader&pg=GBS.PA102}}</ref>
* The [[Catalan number]]s are <math>\frac {\tbinom{2n}n}{n+1}.</math>
* The [[binomial distribution]] in [[statistics]] is <math>\tbinom n k p^k (1-p)^{n-k} \!.</math>
* The formula for a [[Bézier curve]].
 
== Binomial coefficients as polynomials ==
For any nonnegative integer ''k'', the expression <math>\scriptstyle{\binom{t}{k}}</math> can be simplified and defined as a polynomial divided by ''k''!:
 
:<math>\binom{t}{k} =\frac{(t)_k}{k!}=\frac{(t)_k}{(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 number|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 [[Binomial theorem#Newton's generalized binomial theorem|Newton's generalized binomial theorem]].
 
For each ''k'', the polynomial <math>\tbinom{t}{k}</math> can be characterized as the unique degree ''k'' polynomial ''p''(''t'') satisfying ''p''(0) = ''p''(1) = ... = ''p''(''k'' &minus; 1) = 0 and ''p''(''k'') = 1.
 
Its coefficients are expressible in terms of [[Stirling numbers of the first kind]], by definition of the latter:
:<math>\binom{t}{k} = \sum_{i=0}^k \frac{s_{k,i}}{k!} t^i.</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>
 
=== Binomial coefficients as a basis for the space of polynomials ===
Over any [[field (mathematics)|field]] of [[characteristic (algebra)|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>\sum_{k=0}^d a_k \binom{t}{k}</math> of binomial coefficients.  The coefficient ''a''<sub>''k''</sub> is the [[finite difference|''k''<sup>th</sup> difference]] of the sequence ''p''(0), ''p''(1), …, ''p''(''k'').
Explicitly,<ref>This can be seen as a discrete analog of [[Taylor's theorem]]. It is closely related to [[Newton's polynomial]]. Alternating sums of this form may be expressed as the [[Nörlund–Rice integral]].</ref>
{{NumBlk|:|<math>a_k = \sum_{i=0}^k (-1)^{k-i} \binom{k}{i} p(i).</math>|{{EquationRef|4}}}}
 
=== Integer-valued polynomials ===
{{main|Integer-valued polynomial}}
Each polynomial <math>\tbinom{t}{k}</math> is [[integer-valued polynomial|integer-valued]]: it takes integer values at integer inputs.
(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, ({{EquationNote|4}}) 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 3''t''(3''t''&nbsp;+&nbsp;1)/2 can be rewritten as
:<math>9\tbinom{t}{2} + 6 \tbinom{t}{1} + 0\tbinom{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
{{NumBlk|:|<math> \binom{n}{k} = \frac{n}{k} \binom{n-1}{k-1}</math>|{{EquationRef|5}}}}
and, with a little more work,
:<math>\binom {n-1}{k} - \binom{n-1}{k-1} = \frac{n-2k}{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}</math>
 
=== Series involving binomial coefficients ===
The formula
<math> \sum_{k=0}^n \tbinom n k = 2^n</math>
is obtained from ({{EquationNote|2}}) using ''x'' = 1.  This is equivalent to saying that the elements in one row of Pascal's triangle always add up to two raised to an integer power. A combinatorial interpretation of this fact involving [[double counting (proof technique)|double counting]] is given by counting subsets of size 0, size 1, size 2, and so on up to size ''n'' of a set ''S'' of ''n'' elements. Since we count the number of subsets of size ''i'' for 0 ≤ ''i'' ≤ ''n'', this sum must be equal to the number of subsets of ''S'', which is known to be 2<sup>''n''</sup>.  That is, Equation&nbsp;5 is a statement that the [[power set]] for a finite set with ''n'' elements has size 2<sup>''n''</sup>.
More explicitly, consider a bit string with ''n'' digits. This bit string can be used to represent 2<sup>''n''</sup> numbers. Now consider all of the bit strings with no ones in them. There is just one, or rather ''n'' choose 0. Next consider the number of bit strings with just a single one in them. There are ''n'', or rather ''n'' choose 1. Continuing this way we can see that the equation above holds.
 
The formulas
{{NumBlk|:|<math>\sum_{k=0}^n k \tbinom n k = n 2^{n-1}</math>|{{EquationRef|6}}}}
and
<math>\sum_{k=0}^n k^2 \tbinom n k = (n + n^2)2^{n-2}</math>
follow from ({{EquationNote|2}}) after [[derivative|differentiating]]  with respect to ''x'' (twice in the latter)  and then substituting ''x'' = 1.
 
The [[Chu-Vandermonde identity]], which holds for any complex-values ''m'' and ''n'' and any non-negative integer ''k'', is
{{NumBlk|:|<math> \sum_{j=0}^k \tbinom m j \tbinom{n-m}{k-j} = \tbinom n k</math>|{{EquationRef|7}}}}
and can be found by examination of the coefficient of <math>x^k</math> in the expansion of (1&nbsp;+&nbsp;''x'')<sup>''m''</sup>&nbsp;(1&nbsp;+&nbsp;''x'')<sup>''n''&nbsp;−&nbsp;''m''</sup> = (1&nbsp;+&nbsp;''x'')<sup>''n''</sup> using equation ({{EquationNote|2}}).  When ''m''&nbsp;=&nbsp;1, equation ({{EquationNote|7}}) reduces to equation ({{EquationNote|3}}).
 
A similar looking formula, which applies for any integers ''j'', ''k'', and ''n'' satisfying 0&nbsp;≤&nbsp;''j''&nbsp;≤&nbsp;''k''&nbsp;≤&nbsp;''n'', is
{{NumBlk|:|<math>\sum_{m=0}^n \tbinom m j \tbinom {n-m}{k-j}= \tbinom {n+1}{k+1}\,,</math>|{{EquationRef|8}}}}
and can be found by examination of the coefficient of <math>x^{n+1}</math> in the expansion of <math>x\left(\tfrac{x^{j}}{(1-x)^{j+1}}\right) \left(\tfrac{x^{k-j}}{(1-x)^{k-j+1}}\right) = \tfrac{x^{k+1}}{(1-x)^{k+2}}</math>
using <math>\tfrac{x^l}{(1-x)^{l+1}} = \sum_{p=0}^\infty \tbinom p l x^p\,.</math>
When ''j''&nbsp;=&nbsp;''k'', equation ({{EquationNote|8}}) gives
:<math>\sum_{m=0}^n \tbinom m k = \tbinom {n+1}{k+1}\,.</math>
 
From expansion ({{EquationNote|7}}) using ''n''&nbsp;=&nbsp;2''m'', ''k''&nbsp;=&nbsp;''m'', and ({{EquationNote|1}}), one finds
{{NumBlk|:|<math> \sum_{j=0}^m \tbinom m j ^2 = \tbinom {2m} m.</math>|{{EquationRef|9}}}}
 
Let ''F''(''n'') denote the ''n''-th [[Fibonacci number]].
We obtain a formula about the diagonals of Pascal's triangle
<math> \sum_{k=0}^{\lfloor\frac{n}{2}\rfloor} \tbinom {n-k} k = F(n+1).</math>
 
This can be proved by [[mathematical induction|induction]] using ({{EquationNote|3}}) or by [[Zeckendorf's theorem|Zeckendorf's representation]] (Just note that the lhs gives the number of subsets of {''F''(2),...,''F''(''n'')} without consecutive members, which also form all the numbers below ''F''(''n''&nbsp;+&nbsp;1)). A combinatorial proof is given below.
 
Also using ({{EquationNote|3}}) and induction, one can show that
<math> \sum_{j=k}^n \tbinom j k = \tbinom {n+1}{k+1}.</math>
 
Although there is no closed formula for
 
:<math> \sum_{j=0}^k \tbinom n j</math>
 
(unless one resorts to [[Hypergeometric function]]s), one can again use ({{EquationNote|3}}) and induction, to show that for ''k'' = 0, ..., ''n''&nbsp;&minus;&nbsp;1
<math> \sum_{j=0}^k (-1)^j\tbinom n j = (-1)^k\tbinom {n-1}k</math>
 
as well as
<math> \sum_{j=0}^n (-1)^j\tbinom n j = 0</math>
 
[except in the trivial case where ''n'' = 0, where the result is 1 instead] which is itself a special case of the result from the theory of [[finite differences]] that for any polynomial ''P''(''x'') of degree less than ''n'',<ref>{{cite journal|last=Ruiz|first=Sebastian|title=An Algebraic Identity Leading to Wilson's Theorem|journal=The Mathematical Gazette|year=1996|volume=80|issue=489|pages=579–582|url=http://www.jstor.org/stable/3618534}}</ref>
<math> \sum_{j=0}^n (-1)^j\tbinom n j P(j) = 0.</math>
Differentiating ({{EquationNote|2}}) ''k'' times and setting ''x'' = &minus;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'',
 
{{NumBlk|:|<math> \sum_{j=0}^n (-1)^j\tbinom n j P(n-j) = n!a_n</math>|{{EquationRef|10}}}}
 
where <math>a_n</math> is the coefficient of degree ''n'' in ''P''(''x'').
 
More generally for ({{EquationNote|10}}),
 
<math> \sum_{j=0}^n (-1)^j\tbinom n j P(m+(n-j)d) = d^n n! a_n</math>
where ''m'' and ''d'' are complex numbers. This follows immediately applying ({{EquationNote|10}}) to the polynomial ''Q(x):=P(m + dx)'' instead of ''P(x)'', and observing that ''Q(x)'' has still degree less than or equal to ''n'', and that its coefficient of degree ''n'' is ''d<sup>n</sup>a<sub>n</sub>''.
 
The [[infinite series]]
<math>\sum_{j=m}^\infty \frac 1 {\binom j k}=\frac k{(k-1)\binom{m-1}{k-1}}</math>
is convergent for ''k'' ≥ 2. This formula is used in the analysis of the [[German tank problem]].
It is equivalent to the formula for the finite sum
:<math>\sum_{j=m}^{M-1}\frac 1 {k\binom j k}=\frac 1{(k-1)\binom{m-1}{k-1}}-\frac 1{(k-1)\binom{M-1}{k-1}}</math>
which is proved for ''M''>''m'' by [[mathematical induction|induction]] on ''M''.
 
Using ({{EquationNote|9}}) one can derive
<math>\sum_{i=0}^{n}{i\binom{n}{i}^2}=\frac{n}{2}\binom{2n}{n}</math>
and
<math>\sum_{i=0}^n{i^2\binom{n}{i}^2}=n^2 \binom{2n-2}{n-1}.</math>
 
[[Series multisection]] gives the following identity for the sum of binomial coefficients taken with a step ''s'' and offset ''t'' <math>(0\leqslant t<s)</math> as a closed-form sum of ''s'' terms:
: <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>
 
=== Identities with combinatorial proofs ===
Many identities involving binomial coefficients can be proved by [[combinatorial proof|combinatorial means]]. For example, the following identity for nonnegative integers <math>{n} \geq {q}</math> (which reduces to ({{EquationNote|6}}) when ''q'' = 1):
 
:<math>\sum_{k=q}^n \tbinom n k \tbinom k q = 2^{n-q}\tbinom n q</math>
 
can be given a [[double counting (proof technique)|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 parameter, because there are <math>\tbinom n q</math> ways of choosing a set of ''q'' marks and they occur in all subsets that additionally contain some subset of the remaining elements, of which there are <math>2^{n-q}.</math>
 
In the Pascal's rule
:<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''] with the right hand side first grouping them into those that contain element ''n'' and those that do not.
 
The identity ({{EquationNote|9}}) also has a combinatorial proof. The identity reads
 
:<math>\sum_{k=0}^n \tbinom n k ^2 = \tbinom {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\tbinom  n k\tbinom n{n-k} = \tbinom {2n} n.</math>
Now apply ({{EquationNote|5}}) to get the result.
 
The identity ({{EquationRef|9}}),
 
:<math>\sum_{k=0}^{\lfloor\frac{n}{2}\rfloor} \tbinom {n-k} k = F(n+1)</math>
 
has the following combinatorial proof. The number <math>\tbinom{n-k}{k}</math> denotes the number of paths in a two-dimensional lattice from <math>(0,0)</math> to <math>(k,n-k)</math> using steps <math>(0,1)</math> and <math>(1,1)</math>. This is easy to see: there are <math>(n-k)</math> steps in total and one may choose the <math>k</math> <math>(0,1)</math> steps. Now, replace each <math>(1,1)</math> step by a <math>(0,2)</math> step; note that there are exactly <math>k</math>. Then one arrives at point <math>(0,n)</math> using steps <math>(0,1)</math> and <math>(0,2)</math>. Doing this for all <math>k</math> between <math>0</math> and <math>\lfloor\frac{n}{2}\rfloor</math> gives all paths from <math>(0,0)</math> to <math>(0,n)</math> using steps <math>(0,1)</math> and <math>(0,2)</math>. Clearly, there are exactly <math>F(n+1)</math> such paths.
 
==== Sum of coefficients row ====
{{See also|Combination#Number of k-combinations for all k}}
 
The number of ''k''-[[Combination#Number of k-combinations for all k|combinations]] for all ''k'', <math>\sum_{0\leq{k}\leq{n}}\binom nk = 2^n</math>, is the sum of the ''n''th 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 <math>\textstyle m, n \in \mathbb{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 = \left \{
\begin{array}{cc}
(-1)^{m+(n+1)/2} \frac{\pi}{2^{n-1}} \binom{n}{m} & n \text{ odd} \\
0 & \text{otherwise} \\
\end{array} \right.</math>
 
: <math>
\int_{-\pi}^{\pi} \cos((2m-n)x)\sin^n x\ dx = \left \{
\begin{array}{cc}
(-1)^{m+(n+1)/2} \frac{\pi}{2^{n-1}} \binom{n}{m} & n \text{ even} \\
0 & \text{otherwise} \\
\end{array} \right.</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.
 
== Generating functions ==
=== Ordinary generating functions ===
For a fixed ''n'', the [[ordinary generating function]] of the sequence <math>{n\choose 0},\;{n\choose 1},\;{n\choose 2},\;\ldots</math> is:
: <math>\sum_k {n\choose k} x^k = (1+x)^n.</math>
 
For a fixed ''k'',  the ordinary generating function of the sequence <math>{0\choose k},\;{1\choose k},\;{2\choose k},\;\ldots</math> is:
: <math>\sum_{n=k}^\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,k} {n\choose k} x^k y^n = \frac{1}{1-y-xy}.</math>
 
Another bivariate generating function of the binomial coefficients, which is symmetric, is:
: <math>\sum_{n,k} {n+k\choose k} x^k y^n = \frac{1}{1-x-y}.</math>
 
=== Exponential generating function ===
The [[exponential generating function|exponential bivariate generating function]] of the binomial coefficients is:
: <math>\sum_{n,k} \frac{1}{(n+k)!}{n+k\choose k} x^k y^n = e^{x+y}.</math>
 
== Divisibility properties ==
{{main|Kummer's theorem|Lucas' theorem}}
In 1852, [[Ernst Kummer|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''<sup>''c''</sup>, 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''<sup>''j''</sup> is greater than the fractional part of ''n''/''p''<sup>''j''</sup>.  It can be deduced from this that <math>\tbinom n k</math> is divisible by ''n''/[[greatest common divisor|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 ''s''&nbsp;<&nbsp;''p''<sup>''r''</sup>. 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.
 
Another fact:
An integer ''n''&nbsp;≥&nbsp;2 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 0&nbsp;<&nbsp;''k''&nbsp;<&nbsp;''p''
because it is a natural number and the numerator has a prime factor ''p'' but the denominator does not have a prime factor ''p''.
 
When ''n'' is composite, let ''p'' be the smallest prime factor of ''n'' and let ''k''&nbsp;=&nbsp;n/p. Then 0&nbsp;<&nbsp;''p''&nbsp;<&nbsp;''n'' 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 ''k''(''n''&nbsp;&minus;&nbsp;1)(''n''&nbsp;&minus;&nbsp;2)&times;...&times;(''n''&nbsp;&minus;&nbsp;''p''&nbsp;+&nbsp;1) has to be divisible by ''n''&nbsp;=&nbsp;''k''&times;''p'', this can only be the case when (''n''&nbsp;&minus;&nbsp;1)(''n''&nbsp;&minus;&nbsp;2)&times;...&times;(''n''&nbsp;&minus;&nbsp;''p''&nbsp;+&nbsp;1) is divisible by ''p''. But ''n'' is divisible by ''p'', so ''p'' does not divide ''n''&nbsp;&minus;&nbsp;1, ''n''&nbsp;&minus;&nbsp;2, ..., ''n''&nbsp;&minus;&nbsp;''p''&nbsp;+&nbsp;1 and because ''p'' is prime, we know that ''p'' does not divide (''n''&nbsp;&minus;&nbsp;1)(''n''&nbsp;&minus;&nbsp;2)&times;...&times;(''n''&nbsp;&minus;&nbsp;''p''&nbsp;+&nbsp;1) and so the numerator cannot be divisible by ''n''.
 
== Bounds and asymptotic formulas ==
The following bounds for <math>\tbinom n k</math> hold:
 
:<math>\left(\frac{n}{k}\right)^k \le {n \choose k}  \le \frac{n^k}{k!} \le \left(\frac{n\cdot e}{k}\right)^k</math> for 1&nbsp;≤&nbsp;''k''&nbsp;≤&nbsp;''n''.
 
[[Stirling's approximation]] yields the bounds:
 
:<math>\sqrt{n}{2n \choose n} \ge 2^{2n-1}</math> and, in general, <math>\sqrt{n}{mn \choose n} \ge \frac{m^{m(n-1)+1}}{(m-1)^{(m-1)(n-1)}}</math> for ''m''&nbsp;≥&nbsp;2 and ''n''&nbsp;≥&nbsp;1,
 
and the approximation
 
: <math> {2n \choose n} \sim \frac{4^n}{\sqrt{\pi n}}</math> as <math>n\rightarrow\infty\,.</math>
 
For both <math>n</math> and <math>k</math> much larger than 1, [[Stirling's approximation]] also yields the following asymptotic approximation:
 
:<math>\log{n \choose k} \sim n H\left(\frac{k}{n}\right) </math>
 
Where <math>H(\epsilon) = -\epsilon\log(\epsilon) -(1-\epsilon)\log(1-\epsilon)</math> is the [[binary entropy function|binary entropy]] of <math>\epsilon</math>.
 
When <math>n</math> is large and <math>k</math> is much smaller than <math>n</math>, one can also write
 
:<math>{n \choose k} = \frac{n(n-1)\dots(n-k+1)}{k!} \approx \frac{(n-k/2)^k }{ k^k e^{-k} \sqrt{2\pi k}} = \frac{(n/k-0.5)^k e^k }{ \sqrt{2\pi k}}</math>
 
and therefore
 
:<math>\log{n \choose k} \approx k \ln(n/k-0.5) + k - 0.5 \ln(2 \pi k)</math>
 
If more precision is desired, one can approximate <math>\ln{(n(n-1)\dots(n-k+1))}</math> with an integral, obtaining
 
:<math>\log{n \choose k} \approx (n+0.5) \ln\frac{n+0.5}{n-k+0.5} + k \ln \frac{n-k+0.5}{k} - 0.5 \ln(2 \pi k)</math>
 
For <math>n=20</math> and <math>k=10</math>, <math>\log{\tbinom n k} \approx 12.127</math>, and these approximations yield 12.312 and 12.133 respectively.
 
The infinite product formula (cf. [[Gamma_function#Alternative_definitions|Gamma function, alternative definition]])
:<math>(-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{(1+\frac{1}{j})^{-z-1}}{1-\frac{z+1}{j}}</math>
yields the asymptotic formulas
:<math>{z \choose k} \approx \frac{(-1)^k}{\Gamma(-z) k^{z+1}} \qquad  \mathrm{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>{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]].)
 
The sum of binomial coefficients can be bounded by a term exponential in <math>n</math> and the binary entropy of the largest <math>n/k</math> that occurs. More precisely, for <math>n\geq 1</math> and <math>0<\epsilon<\frac{1}{2}</math>, it holds
:<math>\sum_{k=0}^{\lfloor\epsilon n\rfloor} {n \choose k} \leq 2^{H(\epsilon) \cdot n},</math>
where <math>H(\epsilon)=-\epsilon\cdot \log \epsilon - (1-\epsilon)\cdot \log (1-\epsilon)</math> is the [[binary entropy function|binary entropy]] of <math>\epsilon</math>.<ref>see e.g. {{harvtxt |Flum|Grohe|2006| p=427}}</ref>
 
A simple and rough upper bound for the sum of binomial coefficients is given by the formula below (not difficult to prove)
:<math>\sum_{i=0}^k {n \choose i} \leq (n+1)^k</math>
 
==Generalizations==
===Generalization to multinomials===
{{main|Multinomial theorem}}
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 (''x''+''y'')<sup>''n''</sup>, 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<sub>i</sub>'' elements, where ''i'' is the index of the container.
 
Multinomial coefficients have many properties similar to these 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 [[Taylor series|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 ''n''=½ ===
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\geqslant0}{\tbinom{1/2}{k}}x^k.</math>
 
=== Identity for the product of binomial coefficients ===
One can express the product of binomial coefficients as a linear combination of binomial coefficients:
 
:<math> {z \choose m} {z\choose n} = \sum_{k=0}^m {m+n-k\choose k,m-k,n-k} {z\choose m+n-k}</math>
 
where the connection coefficients are [[Multinomial theorem|multinomial coefficients]]. In terms of labelled combinatorial objects, the connection coefficients represent the number of ways to assign ''m+n-k'' labels to a pair of labelled combinatorial objects&mdash;of weight ''m'' and ''n'' respectively&mdash;that have had their first ''k'' labels identified, or glued together to get a new labelled combinatorial object of weight ''m+n-k''. (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 factorial]]s are to ordinary generating series.
 
=== Partial Fraction Decomposition ===
The [[partial fraction decomposition]] of the reciprocal is given by
:<math>\frac{1}{{z \choose n}}= \sum_{i=0}^{n-1} (-1)^{n-1-i} {n \choose i} \frac{n-i}{z-i}, </math> and <math>\frac{1}{{z+n \choose n}}= \sum_{i=1}^n (-1)^{i-1} {n \choose i} \frac{i}{z+i}.</math>
 
=== Newton's binomial series ===
{{main|binomial series}}
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]] (1 + ''z'') ''f'''(''z'') = α ''f''(''z'').
 
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 ===
{{main|Multichoose#Counting multisets}}
Binomial coefficients count subsets of prescribed size from a given set.  A related combinatorial problem is to count [[multiset]]s 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#Counting multisets|multiset coefficients]]'';<ref>{{citation
| last = Munarini | first = Emanuele
| doi = 10.2298/AADM110609014M
| issue = 2
| journal = Applicable Analysis and Discrete Mathematics
| mr = 2867317
| pages = 176–200
| title = Riordan matrices and sums of harmonic numbers
| volume = 5
| year = 2011}}.</ref> the number of ways to "multichoose" (i.e., choose with replacement) ''k'' items from an ''n'' element set is denoted <math>\left(\!\!\binom n k\!\!\right)</math>.
 
To avoid ambiguity and confusion with ''n''’s main denotation in this article,<br> let ''f'' = ''n'' = ''r'' + (''k'' - 1) and ''r'' = ''f'' - (''k'' - 1).
 
Multiset coefficients may be expressed in terms of binomial coefficients by the rule
:::::<math>\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>(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>{\color{white}^\big|}r^{(k)}=\,r^{\overline k}=\,r\cdot(r+1)\cdot(r+2)\cdot(r+3)\cdots(r+k-1)</math>;
so, for example,
:::::<math>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>\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>\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 ====
If <math>n<0</math>, <math>r=|n|</math> and <math>f=r+k-1</math>, then
 
:<math>\begin{align}\binom{-r}{k} &= \frac{-r\cdot-(r+1)\dots-(r+k-2)\cdot-(r+k-1)}{1\cdot2\cdot3\cdot4\cdot5\cdots k}\\
&=(-1)^k\;\frac{r\cdot(r+1)\cdot(r+2)\cdots(f-2)\cdot(f-1)\cdot f}{1\cdot2\cdot3\cdot4\cdot5\cdots k}\\
&=(-1)^k\binom{f}{k}\\
&=(-1)^k\left(\!\!\binom{f-k+1}{k}\!\!\right)\\
&=(-1)^k\left(\!\!\binom{r}{k}\!\!\right)\;,\end{align}</math>
which extends to all <math>n</math>.
 
In the special case <math>n = -1</math>, this reduces to <math>(-1)^k=\binom{-1}{k}=\left(\!\!\binom{-k}{k}\!\!\right) \,.</math>
 
Therefore, if ''n'' = -4 and ''k'' = 7, then ''r'' = 4 and ''f'' = 10:
:<math>\begin{align}\binom{-4}{6} &= \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>
 
 
thus
: <math>\left(\!\!\binom{r}{k}\!\!\right) = \binom{f}{k}
=\binom{r+k-1}{k}=\binom{-1}{k}\binom{-r}{k} = (-1)^k \binom{-r}{k}.</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(x-y+1,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 {{Harv|Fowler|1996}}. Notably, many binomial identities fail: <math>\textstyle{{n \choose m} = {n \choose n-m}}</math> but <math>\textstyle{{-n \choose m} = {-n \choose -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>(-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 [[Cardinal Number|infinite cardinals]] by defining:
 
:<math>{\alpha \choose \beta} = | \{ B \subseteq A : |B| = \beta \} |</math>
 
where A 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>{\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>{\alpha \choose \alpha} = 2^{\alpha}</math> for any infinite cardinal <math>\alpha</math>.
 
==Binomial coefficient in programming languages==
The notation <math> {n \choose k} </math> is convenient in handwriting but inconvenient for [[typewriter]]s and [[computer terminal]]s. Many [[programming language]]s do not offer a standard [[subroutine]] for computing the binomial coefficient, but for example both the [[APL programming language]] and the (related) [[J programming language]] use the exclamation mark:  k ! n .
 
Naive implementations of the factorial formula, such as the following snippet in [[Python (programming language)|Python]]:
<syntaxhighlight lang="python">
from math import factorial
def binomialCoefficient(n, k):
    return factorial(n) // (factorial(k) * factorial(n - k))
</syntaxhighlight>
 
are very slow and are useless for calculating factorials of very high numbers (in languages such as [[C (programming language)|C]] or [[Java (programming language)|Java]] they suffer from overflow errors because of this reason). A direct implementation of the multiplicative formula works well:
 
<syntaxhighlight lang="python">
def binomialCoefficient(n, k):
    if k < 0 or k > n:
        return 0
    if k == 0 or k == n:
        return 1
    diff = n - k
    if k > diff: # take advantage of symmetry
        k = diff
    c = 1
    for i in range(k):
        c = c * (n - i) / (i + 1)
    return c
</syntaxhighlight>
(In Python, range(k) produces a list from 0 to k-1.)
 
[[Pascal's rule]] provides a recursive definition which can also be implemented in [[Python (programming language)|Python]], although it is less efficient:
<syntaxhighlight lang="python">
def binomialCoefficient(n, k):
    if k < 0 or k > n:
        return 0
    if k > n - k: # take advantage of symmetry
        k = n - k
    if k == 0 or n <= 1:
    return 1
    return binomialCoefficient(n-1, k) + binomialCoefficient(n-1, k-1)
</syntaxhighlight>
 
The example mentioned above can be also written in functional style. The following [[Scheme (programming language)|Scheme]] example uses the recursive definition
:<math>{n \choose k+1} = \frac{n-k}{k+1} {n \choose k} </math>
Rational arithmetic can be easily avoided using integer division
:<math>{n \choose k+1} = \left[(n-k) {n \choose k}\right] \div (k+1) </math>
The following implementation uses all these ideas
<syntaxhighlight lang="scheme">
(define (binomial n k)
;; Helper function to compute C(n,k) via forward recursion
  (define (binomial-iter n k i prev)
    (if (>= i k)
      prev
    (binomial-iter n k (+ i 1) (/ (* (- n i) prev) (+ i 1)))))
;; Use symmetry property C(n,k)=C(n, n-k)
  (if (< k (-  n k))
    (binomial-iter n k 0 1)
    (binomial-iter n (- n k) 0 1)))
</syntaxhighlight>
 
Another way to compute the binomial coefficient when using large numbers is to recognize that
 
:<math>
  {n \choose k} = \frac{n!}{k!\,(n-k)!} = \frac{\Gamma(n+1)}{\Gamma(k+1)\,\Gamma(n-k+1)} = \exp(\ln\Gamma(n+1)-\ln\Gamma(k+1)-\ln\Gamma(n-k+1)),</math>
 
where <math>\ln</math>&thinsp;<math>\Gamma(n)</math> denotes the [[natural logarithm]] of the [[gamma function]] at <math>n</math>. It is a special function that is easily computed and is standard in some programming languages such as using ''log_gamma'' in [[Maxima (software)|Maxima]], ''LogGamma'' in [[Mathematica]], ''gammaln'' in [[MATLAB]], or ''lgamma'' in [[R_(programming_language)|R]]. Roundoff error may cause the returned value to not be an integer.
 
==See also==
* [[Binomial transform]]
* [[Central binomial coefficient]]
* [[Kummer's theorem]] on prime-power divisors of binomial coefficients
* [[List of factorial and binomial topics]]
* [[Lucas' theorem]]
* [[Multiplicities of entries in Pascal's triangle]]
* [[Star of David theorem]]
* [[Sun's curious identity]]
* [[Table of Newtonian series]]
 
== Notes ==
{{Reflist|colwidth=25em}}
 
== References ==
{{refbegin|colwidth=25em}}
* [[Arthur T. Benjamin|Benjamin, Arthur T.]]; Quinn, Jennifer (2003). [https://www.maa.org/EbusPPRO/Bookstore/ProductDetail/tabid/170/Default.aspx?ProductId=675 Proofs that Really Count: The Art of Combinatorial Proof ], Mathematical Association of America.
* {{cite book | first=Victor | last=Bryant | authorlink=Victor Bryant
  | title=Aspects of combinatorics | publisher= Cambridge University Press | year=1993 | isbn=0-521-41974-3
  }}
* {{cite book
| last=Flum  | first=Jörg
| last2=Grohe | first2=Martin
| title = Parameterized Complexity Theory | year = 2006 | publisher = Springer
| url = http://www.springer.com/east/home/generic/search/results?SGWID=5-40109-22-141358322-0
| isbn = 978-3-540-29952-3 |ref=harv
}}
* {{Cite journal | doi = 10.2307/2975209 | title = The Binomial Coefficient Function | first =  David | last = Fowler | authorlink = David Fowler (mathematician) | journal = [[The American Mathematical Monthly]] | volume = 103 |date=January 1996 | pages = 1–17 | issue = 1 | publisher = Mathematical Association of America | ref = harv | postscript = <!--None--> | jstor = 2975209 }}
* {{cite journal |first=P. |last=Goetgheluck |title=Computing Binomial Coefficients |journal=American Math. Monthly |volume=94 |year=1987 |pages=360-365 |doi=10.2307/2323099}}
* {{cite book | first=Ronald L.  | last=Graham    | authorlink= Ronald Graham
            | first2=Donald E. | last2=Knuth    | author2-link= Donald Knuth
            | first3=Oren      | last3=Patashnik | author3-link= Oren Patashnik
  | title=Concrete Mathematics | publisher=Addison-Wesley | year=1994 | edition=Second | isbn= 0-201-55802-5 | pages=153–256 | ref=harv
  }}
* {{cite book |first=Nicholas J. |last=Higham |authorlink=Nicholas J. Higham
  |title=Handbook of writing for the mathematical sciences
  |publisher=[[Society for Industrial and Applied Mathematics|SIAM]] |isbn=0-89871-420-6 |year=1998|page=25|ref=harv}}
* {{cite book | first=Donald E.| last=Knuth | authorlink=Donald Knuth
  | title=The Art of Computer Programming, Volume 1: ''Fundamental Algorithms'' | edition=Third | publisher=Addison-Wesley
  | year=1997 | isbn= 0-201-89683-4 | pages=52–74 | ref=harv
  }}
* {{cite journal | first=David | last=Singmaster | authorlink=David Singmaster
  | title=Notes on binomial coefficients. III. Any integer divides almost all binomial coefficients
  | journal=Journal of the London Mathematical Society | volume=8 | issue=3 | year=1974 | pages=555–560
  | doi=10.1112/jlms/s2-8.3.555 | ref=harv
  }}
* {{cite book |first=G. E. |last=Shilov |title=Linear algebra |publisher=Dover Publications |year=1977 |isbn=978-0-486-63518-7|ref=harv}}
{{refend}}
 
==External links==
* {{springer|title=Binomial coefficients|id=p/b016410}}
* [http://www.elektro-energetika.cz/calculations/komb.php?language=english Calculation of Binomial Coefficient]
* {{cite journal |author=[[Andrew Granville]] |url=http://www.cecm.sfu.ca/organics/papers/granville/Binomial/toppage.html |title=Arithmetic Properties of Binomial Coefficients I. Binomial coefficients modulo prime powers |journal=CMS Conf. Proc |volume=20 |pages=151-162 |year=1997}}
 
{{PlanetMath attribution
| id  =  273 | title = Binomial Coefficient
| id2 = 4074 | title2 = Bounds for binomial coefficients
| id3 = 6744 | title3 = Proof that C(n,k) is an integer
| id4 = 6309 | title4 = Generalized binomial coefficients
}}
 
{{DEFAULTSORT:Binomial Coefficient}}
[[Category:Combinatorics]]
[[Category:Factorial and binomial topics]]
[[Category:Integer sequences]]
[[Category:Triangles of numbers]]
 
{{Link GA|es}}

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