# Binomial coefficient

The binomial coefficients can be arranged to form Pascal's triangle.
Visualisation of binomial expansion up to the 4th power

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 $\tbinom nk$. 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 $\tfrac{n!}{k!\,(n-k)!}$. 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, especially in combinatorics. $\tbinom nk$ is often read aloud as "n choose k", because there are $\tbinom nk$ ways to choose k elements from a set of n elements. The properties of binomial coefficients have led to extending the meaning of the symbol $\tbinom nk$ beyond the basic case where n and k are nonnegative integers with kn; such expressions are still called binomial coefficients.

The notation $\tbinom nk$ 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 $\tbinom nk$ 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

$(x+y)^n=\sum_{k=0}^n\binom nk x^{n-k}y^k$

()

(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 $\tbinom nk$ 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 $\tbinom nk$, while the number of ways to write $k=a_1+a_2+\cdots+a_n$ where every ai is a nonnegative integer is given by $\tbinom{n+k-1}{n-1}$. 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 $\tbinom nk$ without actually expanding a binomial power or counting k-combinations.

### Recursive formula

One method uses the recursive, purely additive, formula

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

with initial/boundary values

$\binom n0 = \binom nn = 1 \quad \text{for all integers } n\ge0,$

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 $\tbinom nk$ = 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

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

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

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

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)

$\binom nk = \binom n{n-k} \quad \text{for }\ 0\leq k\leq n,$

(1)

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

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

### 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:

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

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

$(1+X)^\alpha = \sum_{k=0}^\infty {\alpha \choose k} X^k.$

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

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

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.
Main articles: Pascal's triangle and Pascal's rule

Pascal's rule is the important recurrence relation

${n \choose k} + {n \choose k+1} = {n+1 \choose k+1},$

(3)

which can be used to prove by mathematical induction that $\tbinom n k$ 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 $\tbinom n k$ 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 (3) above.

## Combinatorics and statistics

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

• There are $\tbinom n k$ ways to choose k elements from a set of n elements. See Combination.
• There are $\tbinom {n+k-1}k$ ways to choose k elements from a set of n elements if repetitions are allowed. See Multiset.
• There are $\tbinom {n+k} k$ strings containing k ones and n zeros.
• There are $\tbinom {n+1} k$ strings consisting of k ones and n zeros such that no two ones are adjacent.[5]
• The Catalan numbers are $\tfrac{1}{n+1}\tbinom{2n}{n}.$
• The binomial distribution in statistics is $\tbinom n k p^k (1-p)^{n-k} \!.$

## Binomial coefficients as polynomials

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

$\binom{t}{k} =\frac{(t)_k}{k!}=\frac{(t)_k}{(k)_k}= \frac{t(t-1)(t-2)\cdots(t-k+1)}{k(k-1)(k-2)\cdots2 \cdot 1};\,\!$

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 $\tbinom{t}{k}$ 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:

$\binom{t}{k} = \sum_{i=0}^k \left[{ k \atop i} \right]\frac{t^i}{k!}.$

The derivative of $\tbinom{t}{k}$ can be calculated by logarithmic differentiation:

$\frac{\mathrm{d}}{\mathrm{d}t} \binom{t}{k} = \binom{t}{k} \sum_{i=0}^{k-1} \frac{1}{t-i}\,.$

### 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 $\sum_{k=0}^d a_k \binom{t}{k}$ of binomial coefficients. The coefficient ak is the kth difference of the sequence p(0), p(1), …, p(k). Explicitly,[6]

$a_k = \sum_{i=0}^k (-1)^{k-i} \binom{k}{i} p(i).$

(4)

### Integer-valued polynomials

Each polynomial $\tbinom{t}{k}$ 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, (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 3t(3t + 1)/2 can be rewritten as

$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

$\binom{n}{k} = \frac{n}{k} \binom{n-1}{k-1}$

(5)

and, with a little more work,

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

Moreover, the following may be useful:

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

For constant n, we have the following recurrence:

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

### Series involving binomial coefficients

The formula

$\sum_{k=0}^n \tbinom n k = 2^n$

(∗∗)

is obtained from () 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, (∗∗) 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

$\sum_{k=0}^n k \tbinom n k = n 2^{n-1}$

(6)

and $\sum_{k=0}^n k^2 \tbinom n k = (n + n^2)2^{n-2}$ follow from (2) 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

$\sum_{j=0}^k \tbinom m j \tbinom{n-m}{k-j} = \tbinom n k$

(7)

and can be found by examination of the coefficient of $x^k$ in the expansion of (1 + x)m (1 + x)n − m = (1 + x)n using equation (2). When m = 1, equation (7) reduces to equation (3).

A similar looking formula, which applies for any integers j, k, and n satisfying 0 ≤ j ≤ k ≤ n, is

$\sum_{m=0}^n \tbinom m j \tbinom {n-m}{k-j}= \tbinom {n+1}{k+1}\,,$

(8)

and can be found by examination of the coefficient of $x^{n+1}$ in the expansion of $x\left(\tfrac{x^{j}}{(1-x)^{j+1}}\right) \left(\tfrac{x^{k-j}}{(1-x)^{k-j+1}}\right) = \tfrac{x^{k+1}}{(1-x)^{k+2}}$ using $\tfrac{x^l}{(1-x)^{l+1}} = \sum_{p=0}^\infty \tbinom p l x^p\,.$ When j = k, equation (8) gives

$\sum_{m=0}^n \tbinom m k = \tbinom {n+1}{k+1}\,.$

From expansion (7) using n = 2m, k = m, and (1), one finds

$\sum_{j=0}^m \tbinom m j ^2 = \tbinom {2m} m.$

(9)

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

This can be proved by induction using (3) 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 (8) with j=k-1 is $\sum_{j=k}^n (n+1-j) \tbinom {j-1} {k-1} = \tbinom {n+1}{k+1}.$

Although there is no closed formula for

$\sum_{j=0}^k \tbinom n j$

(unless one resorts to Hypergeometric functions),[7] one can again use (3) and induction, to show that for k = 0, ..., n − 1 $\sum_{j=0}^k (-1)^j\tbinom n j = (-1)^k\tbinom {n-1}k$

as well as[8] $\sum_{j=0}^n (-1)^j\tbinom n j = 0$

[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,[9] $\sum_{j=0}^n (-1)^j\tbinom n j P(j) = 0.$ Differentiating (2) k times and setting x = −1 yields this for $P(x)=x(x-1)\cdots(x-k+1)$, 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,

$\sum_{j=0}^n (-1)^j\tbinom n j P(n-j) = n!a_n$

(10)

where $a_n$ is the coefficient of degree n in P(x).

More generally for (10),

$\sum_{j=0}^n (-1)^j\tbinom n j P(m+(n-j)d) = d^n n! a_n$ where m and d are complex numbers. This follows immediately applying (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 dnan.

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

Using (9) one can derive $\sum_{i=0}^{n}{i\binom{n}{i}^2}=\frac{n}{2}\binom{2n}{n}$ and $\sum_{i=0}^n{i^2\binom{n}{i}^2}=n^2 \binom{2n-2}{n-1}.$

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

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

### Identities with combinatorial proofs

Many identities involving binomial coefficients can be proved by combinatorial means. For example, the following identity for nonnegative integers ${n} \geq {q}$ (which reduces to (6) when q = 1):

$\sum_{k=q}^n \tbinom n k \tbinom k q = 2^{n-q}\tbinom n q$

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 $\tbinom n q$ 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 $2^{n-q}.$

In the Pascal's rule

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

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 (9) also has a combinatorial proof. The identity reads

$\sum_{k=0}^n \tbinom n k ^2 = \tbinom {2n} n.$

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

$\sum_{k=0}^n\tbinom n k\tbinom n{n-k} = \tbinom {2n} n.$

Now apply (5) to get the result.

The identity (9),

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

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

#### Sum of coefficients row

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

### Dixon's identity

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

or, more generally,

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

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

### Continuous identities

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

For $\textstyle m, n \in \mathbb{N}$

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

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

$\sum_k {n\choose k} x^k = (1+x)^n.$

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

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

The bivariate generating function of the binomial coefficients is:

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

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

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

### Exponential generating function

The exponential bivariate generating function of the binomial coefficients is:

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

## Divisibility properties

Main articles: Kummer's theorem and 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 $\tbinom{m+n}{m}$ 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 $\tbinom n k$ 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 $\tbinom n k$ is divisible by n/gcd(n,k). In particular therefore it follows that p divides $\tbinom{p^r}{s}$ 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 $\tbinom{9}{6}$.

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

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

Since the number of binomial coefficients $\tbinom n k$ 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

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

are divisible by n.

Proof: When p is prime, p divides

$\binom p k = \frac{p \cdot (p-1) \cdots (p-k+1)}{k \cdot (k-1) \cdots 1}$ 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

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

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 $\tbinom n k$ hold:

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

Stirling's approximation yields the bounds:

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

and the approximation

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

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

$\log_2{n \choose k} \sim n H\left(\frac{k}{n}\right)$

where $H(\epsilon) = -\epsilon\log_2(\epsilon) -(1-\epsilon)\log_2(1-\epsilon)$ is the binary entropy of $\epsilon$. More precisely, for all integers $n \geq k \geq 1$ with $\epsilon \doteq k/n \leq 1/2$, we can estimate the sum of the first $k+1$ binomial coefficients as follows:[10]

$\frac{1}{\sqrt{8n\epsilon(1-\epsilon)}} \cdot 2^{H(\epsilon) \cdot n} \leq \sum_{i=0}^{k} \binom{n}{i} \leq 2^{H(\epsilon) \cdot n}\,.$

When $n$ is large and $k$ is much smaller than $n$, one can also write

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

and therefore

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

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

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

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

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

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

yields the asymptotic formulas

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

as $k \to \infty$.

This asymptotic behaviour is contained in the approximation

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

as well. (Here $H_k$ is the k-th harmonic number and $\gamma$ is the Euler–Mascheroni constant.)

Further, the asymptotic formula

${\frac {{z+k \choose j}}{{k \choose j}}}\to \left(1-{\frac {j}{k}}\right)^{{-z}}\quad {\text{and}}\quad {\frac {{j \choose j-k}}{{j-z \choose j-k}}}\to \left({\frac {j}{k}}\right)^{z}$

hold true, whenever $k\to\infty$ and $j/k\to x$ for some complex number $x$.

A simple and rough upper bound for the sum of binomial coefficients can be obtained using the binomial theorem:

$\sum_{i=0}^k {n \choose i} \leq \sum_{i=0}^k n^i\cdot 1^{k-i} = (1+n)^k$

## Approximations

For large values of n, the following gives an approximation of the binomial coefficient based on its relation to the normal distribution

$\binom{n}{k} = \frac{2^n}{\sqrt{\frac{1}{2}n \pi }} e^{-\frac{(k-(n/2))^2}{n/2}}\left[1+O\left(\frac{1}{\sqrt{n}}\right)\right].$

This follows from the bounds on the Central Limit Theorem, taking a normal distribution with the same expectation and variance as the binomial distribution (np, np(1-p)), centering the probabilities, by setting p = 1 - p = 1/2, and multiplying both by 2n.

## Generalizations

### Generalization to multinomials

Main article: Multinomial theorem

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

${n\choose k_1,k_2,\ldots,k_r} =\frac{n!}{k_1!k_2!\cdots k_r!}$

where

$\sum_{i=1}^rk_i=n.$

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

$(x_1 + x_2 + \cdots + x_r)^n.\$

The case r = 2 gives binomial coefficients:

${n\choose k_1,k_2}={n\choose k_1, n-k_1}={n\choose k_1}= {n\choose k_2}.$

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:

${n\choose k_1,k_2,\ldots,k_r} ={n-1\choose k_1-1,k_2,\ldots,k_r}+{n-1\choose k_1,k_2-1,\ldots,k_r}+\ldots+{n-1\choose k_1,k_2,\ldots,k_r-1}$

and symmetry:

${n\choose k_1,k_2,\ldots,k_r} ={n\choose k_{\sigma_1},k_{\sigma_2},\ldots,k_{\sigma_r}}$

where $(\sigma_i)$ is a permutation of (1,2,...,r).

### Taylor series

Using Stirling numbers of the first kind the series expansion around any arbitrarily chosen point $z_0$ is

\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}

### Binomial coefficient with n = ½

The definition of the binomial coefficients can be extended to the case where $n$ is real and $k$ is integer.

In particular, the following identity holds for any non-negative integer $k$ :

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

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

$\sqrt{1+x}=\sum_{k\geqslant0}{\tbinom{1/2}{k}}x^k.$

### Identity for the product of binomial coefficients

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

${z \choose m} {z\choose n} = \sum_{k=0}^m {m+n-k\choose k,m-k,n-k} {z\choose m+n-k}$

where the connection coefficients are 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—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 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.

### Partial fraction decomposition

The partial fraction decomposition of the reciprocal is given by

$\frac{1}{{z \choose n}}= \sum_{i=0}^{n-1} (-1)^{n-1-i} {n \choose i} \frac{n-i}{z-i},$ and $\frac{1}{{z+n \choose n}}= \sum_{i=1}^n (-1)^{i-1} {n \choose i} \frac{i}{z+i}.$

### Newton's binomial series

Main article: binomial series

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

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

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

$\frac{1}{(1-z)^{\alpha+1}} = \sum_{n=0}^{\infty}{n+\alpha \choose n}z^n$

where the identity

${n \choose k} = (-1)^k {k-n-1 \choose k}$

is applied.

### Multiset (rising) binomial coefficient

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;[11] the number of ways to "multichoose" (i.e., choose with replacement) k items from an n element set is denoted $\left(\!\!\binom n k\!\!\right)$.

To avoid ambiguity and confusion with n’s main denotation in this article,
let f = n = r + (k - 1) and r = f - (k - 1).

Multiset coefficients may be expressed in terms of binomial coefficients by the rule

$\binom{f}{k}=\left(\!\!\binom{r}{k}\!\!\right)=\binom{r+k-1}{k}.$

One possible alternative characterization of this identity is as follows: We may define the falling factorial as

$(f)_{k}=f^{\underline k}=(f-k+1)\cdots(f-3)\cdot(f-2)\cdot(f-1)\cdot f$,

and the corresponding rising factorial as

${\color{white}{ \big|}}r^{(k)}=\,r^{\overline k}=\,r\cdot(r+1)\cdot(r+2)\cdot(r+3)\cdots(r+k-1)$;

so, for example,

$17\cdot18\cdot19\cdot20\cdot21=(21)_{5}=21^{\underline 5}=17^{\overline 5}=17^{(5)}$.

Then the binomial coefficients may be written as

$\binom{f}{k} = \frac{(f)_{k}}{k!} =\frac{(f-k+1)\cdots(f-2)\cdot(f-1)\cdot f}{1\cdot2\cdot3\cdot4\cdot5\cdots k}$,

while the corresponding multiset coefficient is defined by replacing the falling with the rising factorial:

$\left(\!\!\binom{r}{k}\!\!\right)=\frac{r^{(k)}}{k!}=\frac{r\cdot(r+1)\cdot(r+2)\cdots(r+k-1)}{1\cdot2\cdot3\cdot4\cdot5\cdots k}$.

#### Generalization to negative integers

For any n,

\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}

In particular, binomial coefficients evaluated at negative integers are given by signed multiset coefficients. In the special case $n = -1$, this reduces to $(-1)^k=\binom{-1}{k}=\left(\!\!\binom{-k}{k}\!\!\right) \,.$

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

\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}

### 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

${x \choose y}= \frac{\Gamma(x+1)}{\Gamma(y+1) \Gamma(x-y+1)}= \frac{1}{(x+1) \Beta(x-y+1,y+1)}.$

This definition inherits these following additional properties from $\Gamma$:

${x \choose y}= \frac{\sin (y \pi)}{\sin(x \pi)} {-y-1 \choose -x-1}= \frac{\sin((x-y) \pi)}{\sin (x \pi)} {y-x-1 \choose y};$

moreover,

${x \choose y} \cdot {y \choose x}= \frac{\sin((x-y) \pi)}{(x-y) \pi}.$

The resulting function has been little-studied, apparently first being graphed in (Fowler 1996). Notably, many binomial identities fail: $\textstyle{{n \choose m} = {n \choose n-m}}$ but $\textstyle{{-n \choose m} = {-n \choose -n-m}}$ for n positive (so $-n$ 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 $y=x$), with the behavior for negative x having singularities at negative integer values and a checkerboard of positive and negative regions:

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

### Generalization to q-series

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

### Generalization to infinite cardinals

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

${\alpha \choose \beta} = | \{ B \subseteq A : |B| = \beta \} |$

where A is some set with cardinality $\alpha$. 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 $\alpha$, ${\alpha \choose \beta}$ 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 ${\alpha \choose \alpha} = 2^{\alpha}$ for any infinite cardinal $\alpha$.

## Binomial coefficient in programming languages

The notation ${n \choose k}$ is convenient in handwriting but inconvenient for typewriters and computer terminals. Many programming languages 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:

from math import factorial
def binomialCoefficient(n, k):
return factorial(n) // (factorial(k) * factorial(n - k))


are very slow and are useless for calculating factorials of very high numbers (in languages such as C or Java they suffer from overflow errors because of this reason). A direct implementation of the multiplicative formula works well:

def binomialCoefficient(n, k):
if k < 0 or k > n:
return 0
if k == 0 or k == n:
return 1
k = min(k, n - k) # take advantage of symmetry
c = 1
for i in range(k):
c = c * (n - i) / (i + 1)
return c


(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, although it is less efficient:

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)


The example mentioned above can be also written in functional style. The following Scheme example uses the recursive definition

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

Rational arithmetic can be easily avoided using integer division

${n \choose k+1} = \left[(n-k) {n \choose k}\right] \div (k+1)$

The following implementation uses all these ideas

(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)))


Another way to compute the binomial coefficient when using large numbers is to recognize that

${n \choose k} = \frac{n!}{k!\,(n-k)!} = \frac{\Gamma(n+1)}{\Gamma(k+1)\,\Gamma(n-k+1)} = \exp(\ln\Gamma(n+1)-\ln\Gamma(k+1)-\ln\Gamma(n-k+1)),$

where $\ln$$\Gamma(n)$ denotes the natural logarithm of the gamma function at $n$. It is a special function that is easily computed and is standard in some programming languages such as using log_gamma in Maxima, LogGamma in Mathematica, gammaln in MATLAB, or lgamma in R. Roundoff error may cause the returned value to not be an integer.

## Notes

1. ^
2. ^ Lilavati Section 6, Chapter 4 (see Knuth (1997)).
3. ^
4. ^ See (Graham, Knuth & Patashnik 1994), which also defines $\tbinom n k = 0$ for $k<0$. Alternative generalizations, such as to two real or complex valued arguments using the Gamma function assign nonzero values to $\tbinom n k$ for $k<0$, 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).
5. ^ Muir, Thomas (1902). "Note on Selected Combinations". Proceedings of the Royal Society of Edinburgh.
6. ^ 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.
7. ^ Boardman, Michael (2004), "The Egg-Drop Numbers", Mathematics Magazine 77 (5): 368–372, JSTOR 3219201, MR 1573776, it is well known that there is no closed form (that is, direct formula) for the partial sum of binomial coefficients.
8. ^ see induction developed in eq (7) p. 1389 in Aupetit, Michael (2009), "Nearly homogeneous multi-partitioning with a deterministic generator", Neurocomputing 72 (7-9): 1379–1389, doi:10.1016/j.neucom.2008.12.024, ISSN 0925-2312.
9. ^ Ruiz, Sebastian (1996). "An Algebraic Identity Leading to Wilson's Theorem". The Mathematical Gazette 80 (489): 579–582. doi:10.2307/3618534.
10. ^ see e.g. Ash (1990, p. 121) or Flum & Grohe (2006, p. 427).
11. ^ Munarini, Emanuele (2011), "Riordan matrices and sums of harmonic numbers", Applicable Analysis and Discrete Mathematics 5 (2): 176–200, doi:10.2298/AADM110609014M, MR 2867317.