Binomial coefficient

{{short description|Number of subsets of a given size}} {{redirect|nCk||NCK (disambiguation)}} [[File:Pascal's triangle 5.svg|right|thumb|200px|The binomial coefficients can be arranged to form Pascal's triangle, in which each entry is the sum of the two immediately above.|alt=The first five rows of Pascal's triangle, arranged in a triangular formation. Values: (1, (1, 1), (1, 2, 1), (1, 3, 3, 1), (1, 4, 6, 4, 1), (1, 5, 10, 10, 5, 1))]] [[File:Binomial theorem visualisation.svg|thumb|300px|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 {{math|''n'' ≥ ''k'' ≥ 0}} and is written <math>\tbinom{n}{k}</math> or {{tmath|C(n, k)}}. It is the coefficient of the {{math|''x''''k''}} term in the polynomial expansion of the binomial power {{math|(1 + ''x'')''n''}}; this coefficient can be computed by the multiplicative formula

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

For example, the fourth power of {{math|1 + ''x''}} is <math display="block">\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 {{math|''x''2}} term.

Arranging the numbers <math>\tbinom{n}{0}, \tbinom{n}{1}, \ldots, \tbinom{n}{n}</math> in successive rows for {{math|1= ''n'' = 0, 1, 2, ...}} gives a triangular array called Pascal's triangle, satisfying the recurrence relation <math display="block">\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. In combinatorics the symbol <math>\tbinom{n}{k}</math> is usually read as "{{mvar|n}} choose {{mvar|k}}" because there are <math>\tbinom{n}{k}</math> ways to choose an (unordered) subset of {{mvar|k}} elements from a fixed set of {{mvar|n}} elements. For example, there are <math>\tbinom{4}{2}=6</math> ways to choose {{math|2}} elements from {{math|{{mset|1, 2, 3, 4}}}}, namely {{math|{{mset|1, 2}}}}, {{math|{{mset|1, 3}}}}, {{math|{{mset|1, 4}}}}, {{math|{{mset|2, 3}}}}, {{math|{{mset|2, 4}}}} and {{math|{{mset|3, 4}}}}.

The binomial coefficients can be extended to accept more general families of inputs. When {{mvar|n}} is a nonnegative integer and {{mvar|k}} is an integer such that {{math|''k'' < 0}} or {{math|''k'' > ''n''}}, it is common to define {{tmath|1=\tbinom{n}{k} = 0}}. If {{mvar|k}} is a nonnegative integer and {{mvar|z}} is any complex number, the first multiplicative formula above can be used to define {{tmath|\tbinom{z}{k} }}. Many of the properties of binomial coefficients continue to hold in these more general contexts.

== History and notation == Andreas von Ettingshausen introduced the notation <math>\tbinom nk</math> in 1826,<ref>{{harvtxt|Higham|1998}}</ref> although the numbers were known centuries earlier (see Pascal's triangle). Around 1150, the Indian mathematician Bhaskaracharya gave an exposition of binomial coefficients in his book ''Līlāvatī''.<ref>Lilavati Section 6, Chapter 4 (see {{harvtxt|Knuth|1997}}).</ref>

Alternative notations include {{math|''C''(''n'', ''k'')}}, {{math|{{sub|''n''}}''C''{{sub|''k''}}}}, {{math|{{sup|''n''}}''C''''k''}}, {{math|''C''{{su|p=''k''|b=''n''|lh=0.8}}}},<ref>{{harvnb|Uspensky|1937|p=18}}</ref> {{math|''C''{{su|p=''n''|b=''k''|lh=0.8}}}}, and {{math|''C''{{sub|''n'',''k''}}}}, in all of which the {{mvar|C}} stands for ''combinations'' or ''choices''; the {{mvar|C}} notation means the number of ways to choose ''k'' out of ''n'' objects. Many calculators use variants of the {{nowrap|{{mvar|C}} notation}} because they can represent it on a single-line display. In this form the binomial coefficients are easily compared to the numbers of {{mvar|k}}-permutations of {{mvar|n}}, written as {{math|''P''(''n'', ''k'')}}, etc.

== Definition and interpretations == {| class="wikitable" style="float:right;line-spacing:0.8;margin-left:1ex;" |- ! {{diagonal split header|''n''|''k''}} !! 0 !! 1 !! 2 !! 3 !! 4 !! ⋯ |- ! 0 | 1 || {{silverC|0}} || {{silverC|0}} || {{silverC|0}} || {{silverC|0}} || ⋯ |- ! 1 | 1 || 1 || {{silverC|0}} || {{silverC|0}} || {{silverC|0}} || ⋯ |- ! 2 | 1 || 2 || 1 || {{silverC|0}} || {{silverC|0}} || ⋯ |- ! 3 | 1 || 3 || 3 || 1 || {{silverC|0}} || ⋯ |- ! 4 | 1 || 4 || 6 || 4 || 1 || ⋯ |- ! ⋮ | ⋮ || ⋮ || ⋮ || ⋮ || ⋮ || ⋱ |- | colspan="7"|The first few binomial coefficients on a left-aligned Pascal's triangle |}

For natural numbers (taken to include 0) {{mvar|n}} and {{mvar|k}}, the binomial coefficient <math>\tbinom nk</math> can be defined as the coefficient of the monomial {{math|''X''''k''}} in the expansion of {{math|(1 + ''X'')''n''}}. The same coefficient also occurs (if {{math|''k'' ≤ ''n''}}) in the binomial formula {{NumBlk2|:|<math>(x+y)^n=\sum_{k=0}^n\binom nk x^ky^{n-k}</math>|∗}} (valid for any elements {{mvar|x}}, {{mvar|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 {{mvar|k}} objects can be chosen from among {{mvar|n}} objects; more formally, the number of {{mvar|k}}-element subsets (or {{mvar|k}}-combinations) of an {{mvar|n}}-element set. This number can be seen as equal to that of the first definition, independently of any of the formulas below to compute it: if in each of the {{mvar|n}} factors of the power {{math|(1 + ''X'')''n''}} one temporarily labels the term {{mvar|X}} with an index {{mvar|i}} (running from {{math|1}} to {{mvar|n}}), then each subset of {{mvar|k}} indices gives after expansion a contribution {{math|''X''''k''}}, 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 {{mvar|n}} and {{mvar|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 {{mvar|n}} bits (digits 0 or 1) whose sum is {{mvar|k}} is given by {{tmath|\tbinom nk}}, while the number of ways to write <math>k = a_1 + a_2 + \cdots + a_n</math> where every {{math|''a''''i''}} is a nonnegative integer is given by {{tmath|1= \tbinom{n+k-1}{n-1} }}. Most of these interpretations can be shown to be equivalent to counting {{mvar|k}}-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 {{mvar|k}}-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 {{tmath|1 \le k < n}}, with boundary values <math display="block">\binom n0 = \binom nn = 1</math> for all integers {{math|1=''n'' ≥ 0}}.

The formula follows from considering the set {{math|{{mset|1, 2, 3, ..., ''n''}}}} and counting separately (a) the {{mvar|k}}-element groupings that include a particular set element, say "{{mvar|i}}", in every group (since "{{mvar|i}}" is already chosen to fill one spot in every group, we need only choose {{math|''k'' − 1}} from the remaining {{math|''n'' − 1}}) and (b) all the ''k''-groupings that don't include "{{mvar|i}}"; this enumerates all the possible {{mvar|k}}-combinations of {{mvar|n}} elements. It also follows from tracing the contributions to ''X''''k'' in {{math|(1 + ''X'')''n''−1(1 + ''X'')}}. As there is zero {{math|''X''''n''+1}} or {{math|''X''−1}} in {{math|(1 + ''X'')''n''}}, one might extend the definition beyond the above boundaries to include <math>\tbinom nk = 0</math> when either {{math|''k'' > ''n''}} or {{math|''k'' < 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 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, {{tmath|n^{\underline{k} } }}, is a falling factorial. 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 {{mvar|k}} distinct objects, retaining the order of selection, from a set of {{mvar|n}} objects. The denominator counts the number of distinct sequences that define the same {{mvar|k}}-combination when order is disregarded. This formula can also be stated in a recursive form. Using the "C" notation from above, {{tmath|1=C_{n,k} = C_{n, k-1} \cdot (n-k+1) / k}}, where {{tmath|1=C_{n,0} = 1}}. It is readily derived by evaluating <math>C_{n,k} / C_{n, k-1}</math> and can intuitively be understood as starting at the leftmost coefficient of the {{tmath|n}}-th row of Pascal's triangle, whose value is always {{tmath|1}}, and recursively computing the next coefficient to its right until the {{tmath|k}}-th one is reached.

Due to the symmetry of the binomial coefficients with regard to {{mvar|k}} and {{math|''n'' − ''k''}}, calculation of the above product, as well as the recursive relation, may be optimised by setting its upper limit to the smaller of {{mvar|k}} and {{math|''n'' − ''k''}}.

=== Factorial formula === Finally, 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 {{math|''n''!}} denotes the factorial of {{mvar|n}}. This formula follows from the multiplicative formula above by multiplying numerator and denominator by {{math|(''n'' − ''k'')!}}; as a consequence it involves many factors common to numerator and denominator. It is less practical for explicit computation (in the case that {{mvar|k}} is small and {{mvar|n}} 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) {{NumBlk2|:|<math> \binom nk = \binom n{n-k} \quad \text{for }\ 0\leq k\leq n,</math>|1}} 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 === {{Main|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 {{tmath|k<0}}. Alternative generalizations, such as to two real or complex valued arguments using the Gamma function assign nonzero values to <math>\tbinom n k</math> for {{tmath|k < 0}}, but this causes most binomial coefficient identities to fail, and thus is not widely used by the 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 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: {{NumBlk2|:|<math> (1+X)^\alpha = \sum_{k=0}^\infty \binom{\alpha}{k} X^k.</math>|2}}

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 {{math|''k'' > ''n''}} are zero,<ref>When <math>\alpha = n</math> is a nonnegative integer, <math>\textstyle \binom{n}{k} = 0</math> for <math>k > n</math> because the {{tmath|1=(k = n+1)}}-th factor of the numerator is {{tmath|1=n - (n+1) + 1 = 0}}. Thus, the {{tmath|k}}-th term is a zero product for all {{tmath|k \geq n + 1}}.</ref> 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 log-concave sequence.]]

Pascal's rule is the important recurrence relation {{NumBlk2|:|<math> \binom{n}{k} + \binom{n}{k+1} = \binom{n+1}{k+1},</math>|3}} 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 {{mvar|n}} contains the numbers <math>\tbinom{n}{k}</math> for {{math|1=''k'' = 0, …, ''n''}}. 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 display="block">(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.<ref>{{cite journal|last=Muir|first=Thomas|title=Note on Selected Combinations|journal=Proceedings of the Royal Society of Edinburgh|year=1902|url=https://books.google.com/books?id=EN8vAAAAIAAJ&pg=GBS.PA102}}</ref> * The Catalan numbers are {{tmath|\tfrac{1}{n+1}\tbinom{2n}{n} }}. * The binomial distribution in statistics is {{tmath|\tbinom n k p^k (1-p)^{n-k} }}.

== Binomial coefficients as polynomials == For any nonnegative integer ''k'', the expression <math display="inline">\binom{t}{k}</math> can be written as a polynomial with denominator {{math|''k''!}}: <math display="block">\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 {{math|''p''(''t'')}} satisfying {{math|1=''p''(0) = ''p''(1) = ⋯ = ''p''(''k'' − 1) = 0}} and {{math|1=''p''(''k'') = 1}}.

Its coefficients are expressible in terms of Stirling numbers of the first kind: <math display="block">\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 display="block">\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 {{tmath|t-1}}, but using identities below we can compute the derivative as: <math display="block">\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, because the binomial coefficients consist of one polynomial of each degree. The coefficient ''a''''k'' is the ''k''th 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> {{NumBlk2|:|<math>a_k = \sum_{i=0}^k (-1)^{k-i} \binom{k}{i} p(i).</math>|4}}

=== Integer-valued polynomials === {{main|Integer-valued polynomial}} Each polynomial <math>\tbinom{t}{k}</math> is integer-valued: it has an integer value at all integer inputs {{tmath|t}}. (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 {{math|3''t''(3''t'' + 1) / 2}} can be rewritten as <math display="block">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 {{NumBlk2|:|<math> \binom{n}{k} = \frac{n}{k} \binom{n-1}{k-1}</math>|5}} and, with a little more work, <math display="block">\binom {n-1}{k} - \binom{n-1}{k-1} = \frac{n-2k}{n} \binom{n}{k}.</math>

We can also get <math display="block">\binom {n-1}{k} = \frac{n-k}{n} \binom {n}{k}.</math>

Moreover, the following may be useful: <math display="block">\binom{n}{k}\binom{k}{j} = \binom{n}{j}\binom{n-j}{k-j}=\binom{n}{k-j}\binom{n-k+j}{j}.</math>

For constant ''n'', we have the following recurrence: <math display="block"> \binom{n}{k} = \frac{n-k+1}{k} \binom{n}{k-1}.</math>

To sum up, we have <math display="block">\binom {n}{k} = \binom n{n-k} = \frac{n-k+1}{k} \binom {n}{k-1} = \frac{n}{n-k} \binom {n-1}{k} </math> <math display="block">= \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 {{NumBlk2|:|<math> \sum_{k=0}^n \binom n k = 2^n</math>|∗∗}} says that the elements in the {{mvar|n}}th row of Pascal's triangle always add up to 2 raised to the {{mvar|n}}th power. This is obtained from the binomial theorem ({{EquationNote|∗}}) by setting {{math|1=''x'' = 1}} and {{math|1=''y'' = 1}}. 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 {{NumBlk2|:|<math>\sum_{k=0}^n k \binom{n}{k} = n 2^{n-1}</math>|6}} and <math display="block">\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 {{mvar|x}} (twice for the latter) and then substituting {{math|1=''x'' = ''y'' = 1}}.

The Chu–Vandermonde identity, which holds for any complex values ''m'' and ''n'' and any non-negative integer ''k'', is {{NumBlk2|:|<math> \sum_{j=0}^k \binom m j \binom{n-m}{k-j} = \binom n k</math>|7}} and can be found by examination of the coefficient of <math>x^k</math> in the expansion of {{math|1=(1 + ''x'')''m''(1 + ''x'')''n''−''m'' = (1 + ''x'')''n''}} using equation ({{EquationNote|2}}). When {{math|1=''m'' = 1}}, equation ({{EquationNote|7}}) reduces to equation ({{EquationNote|3}}). In the special case {{math|1=''n'' = 2''m'', ''k'' = ''m''}}, using ({{EquationNote|1}}), the expansion ({{EquationNote|7}}) becomes (as seen in Pascal's triangle at right) {{Image frame|width=395 |content= <math> \begin{array}{c} 1 \\ 1 \qquad 1 \\ 1 \qquad 2 \qquad 1 \\ {\color{blue}1 \qquad 3 \qquad 3 \qquad 1} \\ 1 \qquad 4 \qquad 6 \qquad 4 \qquad 1 \\ 1 \qquad 5 \qquad 10 \qquad 10 \qquad 5 \qquad 1 \\ 1 \qquad 6 \qquad 15 \qquad {\color{red}20} \qquad 15 \qquad 6 \qquad 1 \\ 1 \qquad 7 \qquad 21 \qquad 35 \qquad 35 \qquad 21 \qquad 7 \qquad 1 \end{array} </math> |caption=Pascal's triangle, rows 0 through 7. Equation {{EquationNote|8}} for {{math|1=''m'' = 3}} is illustrated in rows 3 and 6 as {{tmath|1=1^2 + 3^2 + 3^2 + 1^2 = 20}}. }} {{NumBlk2|:|<math> \sum_{j=0}^m \binom{m}{j} ^2 = \binom {2m} m,</math>|8}} 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 {{math|1=0 ≤ ''j'' ≤ ''k'' ≤ ''n''}}, is {{NumBlk2|:|<math>\sum_{m=0}^n \binom m j \binom {n-m}{k-j}= \binom {n+1}{k+1}.</math>|9}} The proof is similar, but uses the binomial series expansion ({{EquationNote|2}}) with negative integer exponents. When {{math|1=''j'' = ''k''}}, equation ({{EquationNote|9}}) gives the hockey-stick identity <math display="block">\sum_{m=k}^n \binom m k = \binom {n+1}{k+1}</math> and its relative <math display="block">\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 display="block"> \sum_{k=0}^{\lfloor n/2\rfloor} \binom {n-k} k = F(n+1).</math> This can be proved by induction using ({{EquationNote|3}}) or by Zeckendorf's representation. A combinatorial proof is given below.

==== Multisections of sums ==== For integers ''s'' and ''t'' such that {{tmath|0\leq t < s}}, series multisection gives the following identity for the sum of binomial coefficients: <math display="block">\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 {{mvar|s}}, these series have particularly nice forms; for example,<ref>{{harvtxt|Gradshteyn|Ryzhik|2014|pp=3–4}}.</ref> <math display="block"> \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 display="block"> \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 display="block"> \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 display="block"> \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 display="block"> \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 display="block"> \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 display="block"> \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 display="block"> \sum_{j=0}^k \binom n j</math> of binomial coefficients,<ref>{{citation | last = Boardman | first = Michael | issue = 5 | journal = Mathematics Magazine | jstor = 3219201 | doi = 10.2307/3219201 | mr = 1573776 | quote = it is well known that there is no closed form (that is, direct formula) for the partial sum of binomial coefficients | pages = 368–372 | title = The Egg-Drop Numbers | volume = 77 | year = 2004}}.</ref> one can again use ({{EquationNote|3}}) and induction to show that for {{math|1=''k'' = 0, …, ''n'' − 1}}, <math display="block"> \sum_{j=0}^k (-1)^j\binom{n}{j} = (-1)^k\binom {n-1}{k},</math> with special case<ref>see induction developed in eq (7) p. 1389 in {{citation | last = Aupetit | first = Michael | journal = Neurocomputing | pages = 1379–1389 | title = Nearly homogeneous multi-partitioning with a deterministic generator | volume = 72 | number = 7–9 | year = 2009 | issn = 0925-2312 | doi = 10.1016/j.neucom.2008.12.024 }}.</ref>

<math display="block">\sum_{j=0}^n (-1)^j\binom n j = 0</math> for {{math|''n'' > 0}}. 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'',<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|doi=10.2307/3618534| jstor=3618534|arxiv=math/0406086|s2cid=125556648 }}</ref> <math display="block">\sum_{j=0}^n (-1)^j\binom n j P(j) = 0.</math> Differentiating ({{EquationNote|2}}) ''k'' times and setting ''x'' = −1 yields this for {{tmath|1=P(x)=x(x-1)\cdots(x-k+1)}}, 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'', {{NumBlk2|:|<math> \sum_{j=0}^n (-1)^j\binom n j P(n-j) = n!a_n</math>|10}} where <math>a_n</math> is the coefficient of degree ''n'' in ''P''(''x'').

More generally for ({{EquationNote|10}}), <math display="block"> \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 ({{EquationNote|10}}) to the polynomial {{tmath|1=Q(x):=P(m + dx)}} instead of {{tmath|P(x)}}, and observing that {{tmath|Q(x)}} still has degree less than or equal to ''n'', and that its coefficient of degree ''n'' is ''dnan''.

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 {{tmath|{n} \geq {q} }}, the identity <math display="block">\sum_{k=q}^n \binom{n}{k} \binom{k}{q} = 2^{n-q}\binom{n}{q}</math> (which reduces to ({{EquationNote|6}}) 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 display="block">\binom{n}{k} = \binom{n-1}{k-1} + \binom{n-1}{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 ({{EquationNote|8}}) also has a combinatorial proof. The identity reads <math display="block">\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 display="block">\sum_{k=0}^n\binom n k\binom n{n-k} = \binom {2n} n.</math> Now apply ({{EquationNote|1}}) to get the result.

If one denotes by {{math|''F''(''i'')}} the sequence of Fibonacci numbers, indexed so that {{math|1=''F''(0) = ''F''(1) = 1}}, 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.<ref>{{harvnb|Benjamin|Quinn|2003|loc=pp. 4−5}}</ref> One may show by induction that {{math|''F''(''n'')}} counts the number of ways that a {{math|''n'' × 1}} strip of squares may be covered by {{math|2 × 1}} and {{math|1 × 1}} tiles. On the other hand, if such a tiling uses exactly {{mvar|k}} of the {{math|2 × 1}} tiles, then it uses {{math|''n'' − 2''k''}} of the {{math|1 × 1}} tiles, and so uses {{math|''n'' − ''k''}} 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 {{mvar|k}} gives the identity.

==== Sum of coefficients row ==== {{See also|Combination#Number of k-combinations for all k}}

The number of ''k''-combinations for all ''k'', {{tmath|1=\sum_{0\leq{k}\leq{n} }\binom nk = 2^n}}, 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 {{tmath|2^n - 1}}, where each digit position is an item from the set of ''n''.

=== Dixon's identity === Dixon's identity is <math display="block">\sum_{k=-a}^{a}(-1)^{k}\binom{2a}{k+a}^3 =\frac{(3a)!}{(a!)^3}</math> or, more generally, <math display="block">\sum_{k=-a}^a(-1)^k\binom{a+b}{a+k} \binom{b+c}{b+k}\binom{c+a}{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 {{tmath|m, n \in \N}}, <math display="block">\int_{-\pi}^{\pi} \cos((2m-n)x)\cos^n(x)\ dx = \frac{\pi}{2^{n-1}} \binom{n}{m}</math> <math display="block"> \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 display="block"> \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 {{tmath|0\leq k \leq n-1}}. 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 display="block"> \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 {{mvar|n}}, the ordinary generating function of the sequence <math>\tbinom n0,\tbinom n1,\tbinom n2,\ldots</math> is <math display="block">\sum_{k=0}^\infty \binom{n}{k} x^k = (1+x)^n.</math>

For a fixed {{mvar|k}}, the ordinary generating function of the sequence {{tmath|\tbinom 0k,\tbinom 1k, \tbinom 2k,\ldots}}, is <math display="block">\sum_{n=0}^\infty \binom{n}{k} y^n = \frac{y^k}{(1-y)^{k+1}}.</math>

The bivariate generating function of the binomial coefficients is <math display="block">\sum_{n=0}^\infty \sum_{k=0}^n \binom{n}{k} x^k y^n = \frac{1}{1-y-xy}.</math>

A symmetric bivariate generating function of the binomial coefficients is <math display="block">\sum_{n=0}^\infty \sum_{k=0}^\infty \binom{n+k}{k} x^k y^n = \frac{1}{1-x-y}.</math> which is the same as the previous generating function after the substitution {{tmath|x\to xy}}.

=== Exponential generating function === A symmetric exponential bivariate generating function of the binomial coefficients is: <math display="block">\sum_{n=0}^\infty \sum_{k=0}^\infty \binom{n+k}{k} \frac{x^k y^n}{(n+k)!} = e^{x+y}.</math>

== Divisibility properties == {{main|Kummer's theorem|Lucas' theorem}} 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''. (For example, <math>\tbinom n k</math> is not divisible by ''p'' if every digit in the base-''p'' representation of ''k'' is less than or equal to the corresponding digit in the base-''p'' representation of ''n''.) 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 {{math|''s'' < ''p''''r''}}. However this is not true of higher powers of ''p'': for example 9 does not divide {{tmath|\tbinom{9}{6} }}.

Any integer divides almost all binomial coefficients.<ref>David Singmaster (1974)</ref> More precisely, fix an integer ''d'' and let ''f''(''N'') denote the number of binomial coefficients <math>\tbinom n k</math> with <math>n < N</math> such that ''d'' divides {{tmath|\tbinom n k}}. Then <math display="block"> \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:<ref name="Farhi2007">{{cite journal |last=Farhi |first=Bakir |title=Nontrivial lower bounds for the least common multiple of some finite sequence of integers |journal=Journal of Number Theory |volume=125 |issue=2 |year=2007 |pages=393–411 |doi=10.1016/j.jnt.2006.10.017 |arxiv=0803.0290 |s2cid=115167580 }}</ref> <math display="block">\binom{n+k}k</math> divides {{tmath|\frac{\operatorname{lcm}(n,n+1,\ldots,n+k)}n}}. <math display="block">\binom{n+k}k</math> is a multiple of {{tmath|\frac{\operatorname{lcm}(n,n+1,\ldots,n+k)}{n\cdot \operatorname{lcm}(\binom{k}0,\binom{k}1,\ldots,\binom{k}k)} }}.

Another fact: An integer {{math|''n'' ≥ 2}} is prime if and only if all the intermediate binomial coefficients <math display="block"> \binom n 1, \binom n 2, \ldots, \binom n{n-1} </math> are divisible by ''n''.

Proof: When ''p'' is prime, ''p'' divides <math display="block"> \binom p k = \frac{p \cdot (p-1) \cdots (p-k+1)}{k \cdot (k-1) \cdots 1} </math> for all {{math|1=0 < ''k'' < ''p''}} 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 {{math|1=''k'' = ''n''/''p''}}. Then {{math|1=0 < ''p'' < ''n''}} and <math display="block"> \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 {{math|1=''k''(''n'' − 1)(''n'' − 2)⋯(''n'' − ''p'' + 1)}} has to be divisible by {{math|1=''n'' = ''k''×''p''}}, this can only be the case when {{math|1=(''n'' − 1)(''n'' − 2)⋯(''n'' − ''p'' + 1)}} is divisible by ''p''. But ''n'' is divisible by ''p'', so ''p'' does not divide {{math|1=''n'' − 1, ''n'' − 2, …, ''n'' − ''p'' + 1}} and because ''p'' is prime, we know that ''p'' does not divide {{math|1=(''n'' − 1)(''n'' − 2)⋯(''n'' − ''p'' + 1)}} 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 {{math|1 ≤ ''k'' ≤ ''n''}}: <math display="block">\frac{n^k}{k^k} \le \binom{n}{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"> \binom{n}{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 {{tmath| \geq \frac{n}{k} }}. A similar argument can be made to show the second inequality. The final strict inequality is equivalent to {{tmath|e^k > k^k / k!}}, that is clear since the RHS is a term of the exponential series {{tmath|1= e^k = \sum_{j=0}^\infty k^j/j! }}.

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.<ref name="Farhi2007" />

The following bounds are useful in information theory:<ref name=cover2006>{{cite book |author1=Thomas M. Cover |author2=Joy A. Thomas |title=Elements of Information Theory |date=18 July 2006 |publisher=Wiley |location=Hoboken, New Jersey |isbn=0-471-24195-4}}</ref>{{rp|353}} <math display="block"> \frac{1}{n+1} 2^{nH(k/n)} \leq \binom{n}{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 \binom{n}{k} \leq \sqrt{\frac{n}{2\pi k(n-k)}} 2^{nH(k/n)}</math> for all {{tmath|1 \leq k \leq n-1}}.<ref name=macwilliams1977>{{cite book |author1=F. J. MacWilliams |author2=N. J. A. Sloane |title=The Theory of Error-Correcting Codes |volume=16 |edition=3rd |year=1981 |publisher=North-Holland |isbn=0-444-85009-0}}</ref>{{rp|309}}

=== 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">\binom{n}{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"> \binom{2n}{n} \sim \frac{2^{2n}}{\sqrt{n\pi }}</math> and {{tmath|\sqrt{n}\binom{2n}{n} \ge 2^{2n-1} }}. More generally, for {{math|''m'' ≥ 2}} and {{math|''n'' ≥ 1}} (again, by applying Stirling's formula to the factorials in the binomial coefficient), <math display="block">\sqrt{n}\binom{mn}{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 {{tmath| \binom{n}{k} }}. 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'' − 2''k''.<ref>{{cite book|title=Asymptopia|last1=Spencer|first1=Joel|last2=Florescu|first2=Laura|date=2014| publisher=AMS|isbn=978-1-4704-0904-3|series=Student mathematical library|volume=71|page=66| oclc=865574788|author1-link=Joel Spencer}}</ref>

=== {{mvar|n}} much larger than {{mvar|k}} === If {{mvar|n}} is large and {{mvar|k}} is {{math|''o''(''n'')}} (that is, if {{math| ''k''/''n'' → 0}}), 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 {{mvar|o}} is the little o notation.<ref>{{cite book|title=Asymptopia|last1=Spencer|first1=Joel| last2=Florescu|first2=Laura|date=2014|publisher=AMS|isbn=978-1-4704-0904-3|series=Student mathematical library|volume=71|page=59|oclc=865574788|author1-link=Joel Spencer}}</ref>

=== Sums of binomial coefficients === A simple upper bound for the sum of binomial coefficients can be obtained using a rough estimate for the multiplicative formula for <math>\binom mi</math> and then the binomial theorem: <math display="block"> \sum_{i=0}^k \binom{n}{i} \leq \sum_{i=0}^k n^i \leq \sum_{i=0}^k \binom ki \,n^i\cdot 1^{k-i} =(n+1)^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 {{tmath|\varepsilon \doteq k/n \leq 1/2}}.<ref>see e.g. {{harvtxt|Ash|1990|p=121}} or {{harvtxt|Flum|Grohe|2006|p=427}}.</ref>

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

This asymptotic behaviour is contained in the approximation <math display="block">\binom{z+k}{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">\frac{\binom{z+k}{j}}{\binom{k}{j}}\to \left(1-\frac{j}{k}\right)^{-z}\quad\text{and}\quad \frac{\binom{j}{j-k}}{\binom{j-z}{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 {{tmath|x}}.

== Generalizations == === Generalization to multinomials === {{main|Multinomial theorem}} Binomial coefficients can be generalized to '''multinomial coefficients''' defined to be the number: <math display="block">\binom{n}{k_1,k_2,\ldots,k_r} =\frac{n!}{k_1!k_2!\cdots k_r!}</math> where <math display="block">\sum_{i=1}^rk_i=n.</math>

While the binomial coefficients represent the coefficients of {{math|(''x'' + ''y'')''n''}}, the multinomial coefficients represent the coefficients of the polynomial <math display="block">(x_1 + x_2 + \cdots + x_r)^n.</math> The case ''r'' = 2 gives binomial coefficients: <math display="block">\binom{n}{k_1,k_2}=\binom{n}{k_1, n-k_1}=\binom{n}{k_1}= \binom{n}{k_2}.</math>

The combinatorial interpretation of multinomial coefficients is distribution of ''n'' distinguishable elements over ''r'' (distinguishable) containers, each containing exactly ''ki'' 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 display="block">\binom{n}{k_1,k_2,\ldots,k_r} =\binom{n-1}{k_1-1,k_2,\ldots,k_r}+\binom{n-1}{k_1,k_2-1,\ldots,k_r}+\ldots+\binom{n-1}{k_1,k_2,\ldots,k_r-1}</math> and symmetry: <math display="block">\binom{n}{k_1,k_2,\ldots,k_r} =\binom{n}{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 display="block">\begin{align} \binom{z}{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 \binom{z_0}{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} \binom{j}{i} \frac{s_{k,j}}{k!}.\end{align}</math>

=== Binomial coefficient with {{math|1=''n'' = 1/2}} === 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 {{tmath|k}}: <math display="block">\binom{1/2}{k}=\binom{2k}{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 display="block">\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 display="block">\binom{z}{m} \binom{z}{n} = \sum_{k=0}^{\min(m,n)} \binom{m + n - k}{k, m - k, n - k} \binom{z}{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 {{math|''m'' + ''n'' − ''k''}} 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 {{math|''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 factorials are to ordinary generating series.

The product of all binomial coefficients in the ''n''th row of the Pascal triangle is given by the formula: <math display="block">\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 display="block">\frac{1}{\binom{z}{n}}= \sum_{i=0}^{n-1} (-1)^{n-1-i} \binom{n}{i} \frac{n-i}{z-i}, \qquad \frac{1}{\binom{z+n}{n}}= \sum_{i=1}^n (-1)^{i-1} \binom{n}{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 display="block"> (1+z)^{\alpha} = \sum_{n=0}^{\infty}\binom{\alpha}{n}z^n = 1+\binom{\alpha}{1}z+\binom{\alpha}{2}z^2+\cdots.</math>

The identity can be obtained by showing that both sides satisfy the differential equation {{math|1=(1 + ''z'') ''f'''(''z'') = ''α'' ''f''(''z'')}}.

The radius of convergence of this series is 1. An alternative expression is <math display="block">\frac{1}{(1-z)^{\alpha+1}} = \sum_{n=0}^{\infty}\binom{n+\alpha}{n}z^n</math> where the identity <math display="block">\binom{n}{k} = (-1)^k \binom{k-n-1}{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 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'';<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| url = http://www.doiserbia.nb.rs/img/doi/1452-8630/2011/1452-86301100014M.pdf }}.</ref> the number of ways to "multichoose" (i.e., choose with replacement) ''k'' items from an ''n'' element set is denoted {{tmath|\left(\!\!\binom n k\!\!\right)}}.

To avoid ambiguity and confusion with ''n'''s main denotation in this article, let {{math|1=''f'' = ''n'' = ''r'' + (''k'' − 1)}} and {{math|1=''r'' = ''f'' − (''k'' − 1)}}.

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'' ==== {{Pascal_triangle_extended.svg}} For any ''n'', <math display="block">\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 {{tmath|1=n = -1}}, this reduces to {{tmath|1=(-1)^k=\binom{-1}{k}=\left(\!\!\binom{-k}{k}\!\!\right) }}.

For example, if ''n'' = −4 and ''k'' = 7, then ''r'' = 4 and ''f'' = 10: <math display="block">\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 display="block">\binom{x}{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 {{tmath|\Gamma}}: <math display="block">\binom{x}{y}= \frac{\sin (y \pi)}{\sin(x \pi)} \binom{-y-1}{-x-1}= \frac{\sin((x-y) \pi)}{\sin (x \pi)} \binom{y-x-1}{y};</math> moreover, <math display="block">\binom{x}{y} \cdot \binom{y}{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 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 {{tmath|1=y=x}}), 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. These coefficients are polynomials in an indeterminate (traditionally denoted ''q'') and have applications to many enumerative problems in combinatorics, such as counting the number of linear subspaces of a vector space over a finite field and counting the number of subsets of {1, 2, ..., ''n''} with certain symmetries (an instance of the cyclic sieving phenomenon).

=== Generalization to infinite cardinals === The definition of the binomial coefficient can be generalized to infinite cardinals by defining: <math display="block">\binom{\alpha}{\beta} = \left| \left\{ B \subseteq A : \left|B\right| = \beta \right\} \right|</math> where {{mvar|A}} is some set with cardinality {{tmath|\alpha}}. 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 {{tmath|\alpha}}, <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">\binom{\alpha}{\alpha} = 2^{\alpha}</math> for any infinite cardinal {{tmath|\alpha}}.

== Notes == {{reflist|25em}}

== References == {{refbegin|colwidth=25em}} * {{cite book | first=Robert B. | last=Ash | title=Information Theory | publisher = Dover Publications, Inc. | orig-year = 1965| year = 1990 | isbn = 0-486-66521-6 }} * {{cite book | title = Proofs that Really Count: The Art of Combinatorial Proof | first1 = Arthur T. | last1 = Benjamin | author1-link = Arthur T. Benjamin | first2 = Jennifer J. | last2 = Quinn | author2-link = Jennifer Quinn | publisher = Mathematical Association of America | series = Dolciani Mathematical Expositions | volume = 27 | year = 2003 | isbn = 978-0-88385-333-7 | title-link = Proofs That Really Count }} * {{cite book | first=Victor | last=Bryant | author-link=Victor Bryant | title=Aspects of combinatorics | publisher= Cambridge University Press | year=1993 | isbn=0-521-41974-3 }} * {{cite book | last1=Flum | first1=Jörg | last2=Grohe | first2=Martin | author2-link=Martin Grohe | title=Parameterized Complexity Theory | year=2006 | publisher=Springer | url=https://www.springer.com/east/home/generic/search/results?SGWID=5-40109-22-141358322-0 | isbn=978-3-540-29952-3 | access-date=2017-08-28 | archive-date=2007-11-18 | archive-url=https://web.archive.org/web/20071118193434/http://www.springer.com/east/home/generic/search/results?SGWID=5-40109-22-141358322-0 | url-status=dead }} * {{Cite journal | doi = 10.2307/2975209 | title = The Binomial Coefficient Function | first = David | last = Fowler | author-link = David Fowler (mathematician) | journal = The American Mathematical Monthly | volume = 103 |date=January 1996 | pages = 1–17 | issue = 1 | publisher = Mathematical Association of America | jstor = 2975209 }} * {{cite journal |first=P. |last=Goetgheluck |title=Computing Binomial Coefficients |journal=American Mathematical Monthly |volume=94 |issue=4 |year=1987 |pages=360–365 |doi=10.2307/2323099|jstor=2323099 }} * {{cite book |author-last1=Graham |author-first1=Ronald L. |author-link1=Ronald L. Graham |author-last2=Knuth |author-first2=Donald E. |author-link2=Donald E. Knuth |author-last3=Patashnik |author-first3=Oren |author-link3=Oren Patashnik |title=Concrete Mathematics - A foundation for computer science |edition=2nd |publisher=Addison-Wesley Professional |location=Reading, MA, USA |date=February 1994 |pages=154–155 |isbn=0-201-55802-5 |mr=1397498}} * {{cite book | last1 = Gradshteyn | first1 = I. S. | last2 = Ryzhik | first2 = I. M. | title =Table of Integrals, Series, and Products | title-link = Gradshteyn and Ryzhik | edition = 8th | year = 2014 | publisher = Academic Press | isbn = 978-0-12-384933-5 }} * {{citation | last1=Grinshpan | first1=A. Z. | title=Weighted inequalities and negative binomials | doi=10.1016/j.aam.2010.04.004 | year=2010 | journal=Advances in Applied Mathematics | volume=45 | issue=4 | pages=564–606 | doi-access=free }} * {{cite book |first=Nicholas J. |last=Higham |author-link=Nicholas Higham |title=Handbook of writing for the mathematical sciences |url=https://archive.org/details/handbookofwritin0000high |url-access=registration |publisher=SIAM |isbn=0-89871-420-6 |year=1998 |page=[https://archive.org/details/handbookofwritin0000high/page/25 25] }} * {{cite book | first=Donald E.| last=Knuth | author-link=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 }} * {{cite journal | first=David | last=Singmaster | author-link=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 }} * {{cite book |first=G. E. |last=Shilov |title=Linear algebra |publisher=Dover Publications |year=1977 |isbn=978-0-486-63518-7}} * {{citation |first=James|last=Uspensky|author-link=J. V. Uspensky |title=Introduction to Mathematical Probability |year=1937 |publisher=McGraw-Hill |url=https://archive.org/details/in.ernet.dli.2015.263184/page/n25/mode/2up }} {{refend}}

== External links == * {{springer|title=Binomial coefficients|id=p/b016410}} * {{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 |author-link=Andrew Granville |access-date=2013-09-03 |archive-url=https://web.archive.org/web/20150923201436/http://www.cecm.sfu.ca/organics/papers/granville/Binomial/toppage.html |archive-date=2015-09-23 |url-status=dead }}

{{PlanetMath attribution | urlname = Binomialcoefficient | title = Binomial Coefficient | urlname2 = UpperandLowerBoundstoBinomialCoefficient | title2 = Upper and lower bounds to binomial coefficient | urlname3 = Nchooserisaninteger | title3 = Binomial coefficient is an integer | urlname4 = Generalizedbinomialcoefficients | title4 = Generalized binomial coefficients }} {{Factorial and binomial topics}}

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