# Binomial coefficient

The binomial coefficients can be arranged to form Pascal's triangle.

In mathematics, binomial coefficients are a family of positive integers that occur as coefficients in the binomial theorem. They are indexed by two nonnegative integers; the binomial coefficient indexed by n and k is usually written ${\displaystyle {\tbinom {n}{k}}}$. It is the coefficient of the x k term in the polynomial expansion of the binomial power (1 + x) n. Under suitable circumstances the value of the coefficient is given by the expression ${\displaystyle {\tfrac {n!}{k!\,(n-k)!}}}$. 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 areas of mathematics other than algebra, notably in combinatorics. For any set containing n elements, the number of distinct k-element subsets of it that can be formed (the k-combinations of its elements) is given by the binomial coefficient ${\displaystyle {\tbinom {n}{k}}}$. Therefore ${\displaystyle {\tbinom {n}{k}}}$ is often read as "n choose k". The properties of binomial coefficients have led to extending the meaning of the symbol ${\displaystyle {\tbinom {n}{k}}}$ beyond the basic case where n and k are nonnegative integers with kn; such expressions are then still called binomial coefficients.

The notation ${\displaystyle {\tbinom {n}{k}}}$ was introduced by Andreas von Ettingshausen in 1826,[1] 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 an exposition of binomial coefficients in his book Lilavati.[2]

Alternative notations include C(n, k), nCk, nCk, Ckn, Cnk,[3] Cn,k in all of which the C stands for combinations or choices. Many calculators use similar variants of the C notation as it can be represented on a single-line display.

## Definition and interpretations

For natural numbers (taken to include 0) n and k, the binomial coefficient ${\displaystyle {\tbinom {n}{k}}}$ can be defined as the coefficient of the monomial Xk in the expansion of (1 + X)n. The same coefficient also occurs (if kn) in the binomial formula Template:NumBlk (valid for any elements x,y of a commutative ring), which explains the name "binomial coefficient".

Another occurrence of this number is in combinatorics, where it gives the number of ways, disregarding order, that k objects can be chosen from among n objects; more formally, the number of k-element subsets (or k-combinations) 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 (1 + X)n 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 Xk, and the coefficient of that monomial in the result will be the number of such subsets. This shows in particular that ${\displaystyle {\tbinom {n}{k}}}$ 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 bits (digits 0 or 1) whose sum is k is given by ${\displaystyle {\tbinom {n}{k}}}$, while the number of ways to write ${\displaystyle k=a_{1}+a_{2}+\cdots +a_{n}}$ where every ai is a nonnegative integer is given by ${\displaystyle {\tbinom {n+k-1}{n-1}}}$. 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 ${\displaystyle {\tbinom {n}{k}}}$ without actually expanding a binomial power or counting k-combinations.

### Recursive formula

One method uses the recursive, purely additive, formula

${\displaystyle {\binom {n}{k}}={\binom {n-1}{k-1}}+{\binom {n-1}{k}}\quad {\text{for all integers }}n,k:1\leq k\leq n-1,}$

with initial/boundary values

${\displaystyle {\binom {n}{0}}={\binom {n}{n}}=1\quad {\text{for all integers }}n\geq 0,}$

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 − 1 from the remaining n − 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 Xk in (1 + X)n−1(1 + X). As there is zero Xn+1 or X−1 in (1 + X)n, one might extend the definition beyond the above boundaries to include ${\displaystyle {\tbinom {n}{k}}}$ = 0 when either k > n or 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

${\displaystyle {\binom {n}{k}}={\frac {n^{\underline {k}}}{k!}}={\frac {n(n-1)(n-2)\cdots (n-(k-1))}{k(k-1)(k-2)\cdots 1}}=\prod _{i=1}^{k}{\frac {n-(k-i)}{i}}=\prod _{i=1}^{k}{\frac {n+1-i}{i}},}$

where the numerator of the first fraction ${\displaystyle n^{\underline {k}}}$ 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:

${\displaystyle {\binom {n}{k}}={\frac {n!}{k!\,(n-k)!}}\quad {\text{for }}\ 0\leq k\leq n,}$

where n! denotes the factorial of n. This formula follows from the multiplicative formula above by multiplying numerator and denominator by (nk)!; 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)

which leads to a more efficient multiplicative computational routine. Using the falling factorial notation,

${\displaystyle {\binom {n}{k}}={\begin{cases}n^{\underline {k}}/k!&{\text{if }}\ k\leq {\frac {n}{2}}\\n^{\underline {n-k}}/(n-k)!&{\text{if }}\ k>{\frac {n}{2}}\end{cases}}.}$

### Generalization and connection to the binomial series

The multiplicative formula allows the definition of binomial coefficients to be extended[4] 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:

${\displaystyle {\binom {\alpha }{k}}={\frac {\alpha ^{\underline {k}}}{k!}}={\frac {\alpha (\alpha -1)(\alpha -2)\cdots (\alpha -k+1)}{k(k-1)(k-2)\cdots 1}}\quad {\text{for }}k\in \mathbb {N} {\text{ and arbitrary }}\alpha .}$

With this definition one has a generalization of the binomial formula (with one of the variables set to 1), which justifies still calling the ${\displaystyle {\tbinom {\alpha }{k}}}$ binomial coefficients:

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 series with constant coefficient equal to 1; the point is that with this definition all identities hold that one expects for exponentiation, notably

${\displaystyle (1+X)^{\alpha }(1+X)^{\beta }=(1+X)^{\alpha +\beta }\quad {\text{and}}\quad ((1+X)^{\alpha })^{\beta }=(1+X)^{\alpha \beta }.}$

If α is a nonnegative integer n, then all terms with k > 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

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.

{{#invoke:main|main}}

Pascal's rule is the important recurrence relation Template:NumBlk which can be used to prove by mathematical induction that ${\displaystyle {\tbinom {n}{k}}}$ 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 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 7 21 35 35 21 7 1 8: 1 8 28 56 70 56 28 8 1

Row number n contains the numbers ${\displaystyle {\tbinom {n}{k}}}$ 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)5 = 1 x5 + 5 x4y + 10 x3y2 + 10 x2y3 + 5 x y4 + 1 y5.

The differences between elements on other diagonals are the elements in the previous diagonal, as a consequence of the recurrence relation (Template:EquationNote) above.

## Combinatorics and statistics

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

## Binomial coefficients as polynomials

For any nonnegative integer k, the expression ${\displaystyle \scriptstyle {\binom {t}{k}}}$ can be simplified and defined as a polynomial divided by k!:

${\displaystyle {\binom {t}{k}}={\frac {(t)_{k}}{k!}}={\frac {(t)_{k}}{(k)_{k}}}={\frac {t(t-1)(t-2)\cdots (t-k+1)}{k(k-1)(k-2)\cdots 2\cdot 1}};\,\!}$

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 ${\displaystyle {\tbinom {t}{k}}}$ can be characterized as the unique degree k polynomial p(t) satisfying p(0) = p(1) = ... = p(k − 1) = 0 and p(k) = 1.

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

${\displaystyle {\binom {t}{k}}=\sum _{i=0}^{k}{\frac {s_{k,i}}{k!}}t^{i}.}$
${\displaystyle {\frac {\mathrm {d} }{\mathrm {d} t}}{\binom {t}{k}}={\binom {t}{k}}\sum _{i=0}^{k-1}{\frac {1}{t-i}}\,.}$

### 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 ${\displaystyle \sum _{k=0}^{d}a_{k}{\binom {t}{k}}}$ of binomial coefficients. The coefficient ak is the kth difference of the sequence p(0), p(1), …, p(k). Explicitly,[6] Template:NumBlk

### Integer-valued polynomials

{{#invoke:main|main}} Each polynomial ${\displaystyle {\tbinom {t}{k}}}$ is 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, (Template:EquationNote) shows that any integer-valued polynomial is an integer linear combination of these binomial coefficient polynomials. More generally, for any subring R of a characteristic 0 field K, a polynomial in K[t] takes values in R at all integers if and only if it is an R-linear combination of binomial coefficient polynomials.

### Example

The integer-valued polynomial 3t(3t + 1)/2 can be rewritten as

${\displaystyle 9{\tbinom {t}{2}}+6{\tbinom {t}{1}}+0{\tbinom {t}{0}}.\ }$

## 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 Template:NumBlk and, with a little more work,

${\displaystyle {\binom {n-1}{k}}-{\binom {n-1}{k-1}}={\frac {n-2k}{n}}{\binom {n}{k}}.}$

Moreover, the following may be useful:

${\displaystyle {\binom {n}{h}}{\binom {n-h}{k}}={\binom {n}{k}}{\binom {n-k}{h}}.}$

For constant n, we have the following recurrence:

${\displaystyle {\binom {n}{k}}={\frac {n+1-k}{k}}{\binom {n}{k-1}}.}$

### Series involving binomial coefficients

The formula Template:NumBlk is obtained from (Template:EquationNote) by setting x = 1 and y = 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 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 ≤ in, this sum must be equal to the number of subsets of S, which is known to be 2n. That is, (Template:EquationNote) is the statement that the power set of a finite set with n elements has size 2n. More explicitly, consider a bit string with n digits. This bit string can be used to represent 2n 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 Template:NumBlk and ${\displaystyle \sum _{k=0}^{n}k^{2}{\tbinom {n}{k}}=(n+n^{2})2^{n-2}}$ follow from (Template:EquationNote) after 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 Template:NumBlk and can be found by examination of the coefficient of ${\displaystyle x^{k}}$ in the expansion of (1 + x)m (1 + x)n − m = (1 + x)n using equation (Template:EquationNote). When m = 1, equation (Template:EquationNote) reduces to equation (Template:EquationNote).

A similar looking formula, which applies for any integers j, k, and n satisfying 0 ≤ j ≤ k ≤ n, is Template:NumBlk and can be found by examination of the coefficient of ${\displaystyle x^{n+1}}$ in the expansion of ${\displaystyle x\left({\tfrac {x^{j}}{(1-x)^{j+1}}}\right)\left({\tfrac {x^{k-j}}{(1-x)^{k-j+1}}}\right)={\tfrac {x^{k+1}}{(1-x)^{k+2}}}}$ using ${\displaystyle {\tfrac {x^{l}}{(1-x)^{l+1}}}=\sum _{p=0}^{\infty }{\tbinom {p}{l}}x^{p}\,.}$ When j = k, equation (Template:EquationNote) gives

${\displaystyle \sum _{m=0}^{n}{\tbinom {m}{k}}={\tbinom {n+1}{k+1}}\,.}$

From expansion (Template:EquationNote) using n = 2m, k = m, and (Template:EquationNote), one finds Template:NumBlk

Let F(n) denote the n-th Fibonacci number. We obtain a formula about the diagonals of Pascal's triangle ${\displaystyle \sum _{k=0}^{\lfloor {\frac {n}{2}}\rfloor }{\tbinom {n-k}{k}}=F(n+1).}$

This can be proved by induction using (Template:EquationNote) or by 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 + 1)). A combinatorial proof is given below.

Another identity that follows from (Template:EquationNote) with j=k-1 is ${\displaystyle \sum _{j=k}^{n}(n+1-j){\tbinom {j-1}{k-1}}={\tbinom {n+1}{k+1}}.}$

Although there is no closed formula for

${\displaystyle \sum _{j=0}^{k}{\tbinom {n}{j}}}$

(unless one resorts to Hypergeometric functions),[7] one can again use (Template:EquationNote) and induction, to show that for k = 0, ..., n − 1 ${\displaystyle \sum _{j=0}^{k}(-1)^{j}{\tbinom {n}{j}}=(-1)^{k}{\tbinom {n-1}{k}}}$

[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,[8] ${\displaystyle \sum _{j=0}^{n}(-1)^{j}{\tbinom {n}{j}}P(j)=0.}$ Differentiating (Template:EquationNote) k times and setting x = −1 yields this for ${\displaystyle P(x)=x(x-1)\cdots (x-k+1)}$, 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,

where ${\displaystyle a_{n}}$ is the coefficient of degree n in P(x).

More generally for (Template:EquationNote),

${\displaystyle \sum _{j=0}^{n}(-1)^{j}{\tbinom {n}{j}}P(m+(n-j)d)=d^{n}n!a_{n}}$ where m and d are complex numbers. This follows immediately applying (Template:EquationNote) 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 dnan.

The series ${\displaystyle {\frac {k-1}{k}}\sum _{j=0}^{\infty }{\frac {1}{\binom {j+x}{k}}}={\frac {1}{\binom {x-1}{k-1}}}}$ is convergent for k ≥ 2. This formula is used in the analysis of the German tank problem. It follows from ${\displaystyle {\frac {k-1}{k}}\sum _{j=0}^{M}{\frac {1}{\binom {j+x}{k}}}={\frac {1}{\binom {x-1}{k-1}}}-{\frac {1}{\binom {M+x}{k-1}}}}$ which is proved by induction on M.

Series multisection gives the following identity for the sum of binomial coefficients taken with a step s and offset t ${\displaystyle (0\leqslant t as a closed-form sum of s terms:

${\displaystyle {\binom {n}{t}}+{\binom {n}{t+s}}+{\binom {n}{t+2s}}+\ldots ={\frac {1}{s}}\sum _{j=0}^{s-1}\left(2\cos {\frac {\pi j}{s}}\right)^{n}\cos {\frac {\pi (n-2t)j}{s}}.}$

### Identities with combinatorial proofs

Many identities involving binomial coefficients can be proved by combinatorial means. For example, the following identity for nonnegative integers ${\displaystyle {n}\geq {q}}$ (which reduces to (Template:EquationNote) when q = 1):

${\displaystyle \sum _{k=q}^{n}{\tbinom {n}{k}}{\tbinom {k}{q}}=2^{n-q}{\tbinom {n}{q}}}$

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 parameter, because there are ${\displaystyle {\tbinom {n}{q}}}$ 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 ${\displaystyle 2^{n-q}.}$

In the Pascal's rule

${\displaystyle {n \choose k}={n-1 \choose k-1}+{n-1 \choose k}}$

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

${\displaystyle \sum _{k=0}^{n}{\tbinom {n}{k}}^{2}={\tbinom {2n}{n}}.}$

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

${\displaystyle \sum _{k=0}^{n}{\tbinom {n}{k}}{\tbinom {n}{n-k}}={\tbinom {2n}{n}}.}$

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

The identity (9),

${\displaystyle \sum _{k=0}^{\lfloor {\frac {n}{2}}\rfloor }{\tbinom {n-k}{k}}=F(n+1)}$

has the following combinatorial proof. The number ${\displaystyle {\tbinom {n-k}{k}}}$ denotes the number of paths in a two-dimensional lattice from ${\displaystyle (0,0)}$ to ${\displaystyle (k,n-k)}$ using steps ${\displaystyle (0,1)}$ and ${\displaystyle (1,1)}$. This is easy to see: there are ${\displaystyle (n-k)}$ steps in total and one may choose the ${\displaystyle k}$ ${\displaystyle (0,1)}$ steps. Now, replace each ${\displaystyle (1,1)}$ step by a ${\displaystyle (0,2)}$ step; note that there are exactly ${\displaystyle k}$. Then one arrives at point ${\displaystyle (0,n)}$ using steps ${\displaystyle (0,1)}$ and ${\displaystyle (0,2)}$. Doing this for all ${\displaystyle k}$ between ${\displaystyle 0}$ and ${\displaystyle \lfloor {\frac {n}{2}}\rfloor }$ gives all paths from ${\displaystyle (0,0)}$ to ${\displaystyle (0,n)}$ using steps ${\displaystyle (0,1)}$ and ${\displaystyle (0,2)}$. Clearly, there are exactly ${\displaystyle F(n+1)}$ such paths.

#### Sum of coefficients row

The number of k-combinations for all k, ${\displaystyle \sum _{0\leq {k}\leq {n}}{\binom {n}{k}}=2^{n}}$, is the sum of the nth row (counting from 0) of the binomial coefficients. These combinations are enumerated by the 1 digits of the set of base 2 numbers counting from 0 to ${\displaystyle 2^{n}-1}$, where each digit position is an item from the set of n.

### Dixon's identity

${\displaystyle \sum _{k=-a}^{a}(-1)^{k}{2a \choose k+a}^{3}={\frac {(3a)!}{(a!)^{3}}}}$

or, more generally,

${\displaystyle \sum _{k=-a}^{a}(-1)^{k}{a+b \choose a+k}{b+c \choose b+k}{c+a \choose c+k}={\frac {(a+b+c)!}{a!\,b!\,c!}}\,,}$

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

### Continuous identities

Certain trigonometric integrals have values expressible in terms of binomial coefficients:

${\displaystyle \int _{-\pi }^{\pi }\cos((2m-n)x)\cos ^{n}x\ dx={\frac {\pi }{2^{n-1}}}{\binom {n}{m}}}$
${\displaystyle \int _{-\pi }^{\pi }\sin((2m-n)x)\sin ^{n}x\ dx=\left\{{\begin{array}{cc}(-1)^{m+(n+1)/2}{\frac {\pi }{2^{n-1}}}{\binom {n}{m}}&n{\text{ odd}}\\0&{\text{otherwise}}\\\end{array}}\right.}$
${\displaystyle \int _{-\pi }^{\pi }\cos((2m-n)x)\sin ^{n}x\ dx=\left\{{\begin{array}{cc}(-1)^{m+(n+1)/2}{\frac {\pi }{2^{n-1}}}{\binom {n}{m}}&n{\text{ even}}\\0&{\text{otherwise}}\\\end{array}}\right.}$

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 ${\displaystyle {n \choose 0},\;{n \choose 1},\;{n \choose 2},\;\ldots }$ is:

${\displaystyle \sum _{k}{n \choose k}x^{k}=(1+x)^{n}.}$

For a fixed k, the ordinary generating function of the sequence ${\displaystyle {0 \choose k},\;{1 \choose k},\;{2 \choose k},\;\ldots }$ is:

${\displaystyle \sum _{n=k}^{\infty }{n \choose k}y^{n}={\frac {y^{k}}{(1-y)^{k+1}}}.}$

The bivariate generating function of the binomial coefficients is:

${\displaystyle \sum _{n,k}{n \choose k}x^{k}y^{n}={\frac {1}{1-y-xy}}.}$

Another bivariate generating function of the binomial coefficients, which is symmetric, is:

${\displaystyle \sum _{n,k}{n+k \choose k}x^{k}y^{n}={\frac {1}{1-x-y}}.}$

### Exponential generating function

The exponential bivariate generating function of the binomial coefficients is:

${\displaystyle \sum _{n,k}{\frac {1}{(n+k)!}}{n+k \choose k}x^{k}y^{n}=e^{x+y}.}$

## Divisibility properties

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

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 ${\displaystyle {\tbinom {n}{k}}}$ with n < N such that d divides ${\displaystyle {\tbinom {n}{k}}}$. Then

${\displaystyle \lim _{N\to \infty }{\frac {f(N)}{N(N+1)/2}}=1.}$

Since the number of binomial coefficients ${\displaystyle {\tbinom {n}{k}}}$ 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 ≥ 2 is prime if and only if all the intermediate binomial coefficients

${\displaystyle {\binom {n}{1}},{\binom {n}{2}},\ldots ,{\binom {n}{n-1}}}$

are divisible by n.

Proof: When p is prime, p divides

${\displaystyle {\binom {p}{k}}={\frac {p\cdot (p-1)\cdots (p-k+1)}{k\cdot (k-1)\cdots 1}}}$ for all 0 < k < 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 = n/p. Then 0 < p < n and

${\displaystyle {\binom {n}{p}}={\frac {n(n-1)(n-2)\cdots (n-p+1)}{p!}}={\frac {k(n-1)(n-2)\cdots (n-p+1)}{(p-1)!}}\not \equiv 0{\pmod {n}}}$

otherwise the numerator k(n − 1)(n − 2)×...×(n − p + 1) has to be divisible by n = k×p, this can only be the case when (n − 1)(n − 2)×...×(n − p + 1) is divisible by p. But n is divisible by p, so p does not divide n − 1, n − 2, ..., n − p + 1 and because p is prime, we know that p does not divide (n − 1)(n − 2)×...×(n − p + 1) and so the numerator cannot be divisible by n.

## Bounds and asymptotic formulas

The following bounds for ${\displaystyle {\tbinom {n}{k}}}$ hold:

${\displaystyle \left({\frac {n}{k}}\right)^{k}\leq {n \choose k}\leq {\frac {n^{k}}{k!}}\leq \left({\frac {n\cdot e}{k}}\right)^{k}}$ for 1 ≤ k ≤ n.

Stirling's approximation yields the bounds:

${\displaystyle {\sqrt {n}}{2n \choose n}\geq 2^{2n-1}}$ and, in general, ${\displaystyle {\sqrt {n}}{mn \choose n}\geq {\frac {m^{m(n-1)+1}}{(m-1)^{(m-1)(n-1)}}}}$ for m ≥ 2 and n ≥ 1,

and the approximation

${\displaystyle {2n \choose n}\sim {\frac {4^{n}}{\sqrt {\pi n}}}}$ as ${\displaystyle n\rightarrow \infty \,.}$

For both ${\displaystyle n}$ and ${\displaystyle k}$ much larger than 1, Stirling's approximation also yields the following asymptotic approximation:

${\displaystyle \log {n \choose k}\sim nH\left({\frac {k}{n}}\right)}$

When ${\displaystyle n}$ is large and ${\displaystyle k}$ is much smaller than ${\displaystyle n}$, one can also write

${\displaystyle {n \choose k}={\frac {n(n-1)\dots (n-k+1)}{k!}}\approx {\frac {(n-k/2)^{k}}{k^{k}e^{-k}{\sqrt {2\pi k}}}}={\frac {(n/k-0.5)^{k}e^{k}}{\sqrt {2\pi k}}}}$

and therefore

${\displaystyle \log {n \choose k}\approx k\ln(n/k-0.5)+k-0.5\ln(2\pi k)}$

If more precision is desired, one can approximate ${\displaystyle \ln {(n(n-1)\dots (n-k+1))}}$ with an integral, obtaining

${\displaystyle \log {n \choose k}\approx (n+0.5)\ln {\frac {n+0.5}{n-k+0.5}}+k\ln {\frac {n-k+0.5}{k}}-0.5\ln(2\pi k)}$

For ${\displaystyle n=20}$ and ${\displaystyle k=10}$, ${\displaystyle \log {\tbinom {n}{k}}\approx 12.127}$, and these approximations yield 12.312 and 12.133 respectively.

The infinite product formula (cf. Gamma function, alternative definition)

${\displaystyle (-1)^{k}{z \choose k}={-z+k-1 \choose k}={\frac {1}{\Gamma (-z)}}{\frac {1}{(k+1)^{z+1}}}\prod _{j=k+1}{\frac {(1+{\frac {1}{j}})^{-z-1}}{1-{\frac {z+1}{j}}}}}$

yields the asymptotic formulas

${\displaystyle {z \choose k}\approx {\frac {(-1)^{k}}{\Gamma (-z)k^{z+1}}}\qquad \mathrm {and} \qquad {z+k \choose k}={\frac {k^{z}}{\Gamma (z+1)}}\left(1+{\frac {z(z+1)}{2k}}+{\mathcal {O}}\left(k^{-2}\right)\right)}$

This asymptotic behaviour is contained in the approximation

${\displaystyle {z+k \choose k}\approx {\frac {e^{z(H_{k}-\gamma )}}{\Gamma (z+1)}}}$

The sum of binomial coefficients can be bounded by a term exponential in ${\displaystyle n}$ and the binary entropy of the largest ${\displaystyle n/k}$ that occurs. More precisely, for ${\displaystyle n\geq 1}$ and ${\displaystyle 0<\epsilon <{\frac {1}{2}}}$, it holds

${\displaystyle \sum _{k=0}^{\lfloor \epsilon n\rfloor }{n \choose k}\leq 2^{H(\epsilon )\cdot n},}$

A simple and rough upper bound for the sum of binomial coefficients is given by the formula below (not difficult to prove)

${\displaystyle \sum _{i=0}^{k}{n \choose i}\leq 1+n^{k}}$

## Generalizations

### Generalization to multinomials

{{#invoke:main|main}} Binomial coefficients can be generalized to multinomial coefficients defined to be the number:

${\displaystyle {n \choose k_{1},k_{2},\ldots ,k_{r}}={\frac {n!}{k_{1}!k_{2}!\cdots k_{r}!}}}$

where

${\displaystyle \sum _{i=1}^{r}k_{i}=n.}$

While the binomial coefficients represent the coefficients of (x+y)n, the multinomial coefficients represent the coefficients of the polynomial

${\displaystyle (x_{1}+x_{2}+\cdots +x_{r})^{n}.\ }$

The case r = 2 gives binomial coefficients:

${\displaystyle {n \choose k_{1},k_{2}}={n \choose k_{1},n-k_{1}}={n \choose k_{1}}={n \choose k_{2}}.}$

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 these of binomial coefficients, for example the recurrence relation:

${\displaystyle {n \choose k_{1},k_{2},\ldots ,k_{r}}={n-1 \choose k_{1}-1,k_{2},\ldots ,k_{r}}+{n-1 \choose k_{1},k_{2}-1,\ldots ,k_{r}}+\ldots +{n-1 \choose k_{1},k_{2},\ldots ,k_{r}-1}}$

and symmetry:

${\displaystyle {n \choose k_{1},k_{2},\ldots ,k_{r}}={n \choose k_{\sigma _{1}},k_{\sigma _{2}},\ldots ,k_{\sigma _{r}}}}$

where ${\displaystyle (\sigma _{i})}$ is a permutation of (1,2,...,r).

### Taylor series

Using Stirling numbers of the first kind the series expansion around any arbitrarily chosen point ${\displaystyle z_{0}}$ is