Binomial coefficient

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

File:Binomial theorem visualisation.svg

Visualisation of binomial expansion up to the 4th power

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

(\binom{n}{k}) = \frac{n \times (n - 1) \times \dots \times (n - k + 1)}{k \times (k - 1) \times \dots \times 1},

which using factorial notation can be compactly expressed as

(\binom{n}{k}) = \frac{n!}{k! (n - k)!} .

For example, the fourth power of Template:Math is

\begin{matrix} (1 + x)^{4} & = (\binom{4}{0}) x^{0} + (\binom{4}{1}) x^{1} + (\binom{4}{2}) x^{2} + (\binom{4}{3}) x^{3} + (\binom{4}{4}) x^{4} \\ = 1 + 4 x + 6 x^{2} + 4 x^{3} + x^{4}, \end{matrix}

and the binomial coefficient $(\binom{4}{2}) = \frac{4 \times 3}{2 \times 1} = \frac{4!}{2! 2!} = 6$ is the coefficient of the Template:Math term.

Arranging the numbers $(\binom{n}{0}), (\binom{n}{1}), \dots, (\binom{n}{n})$ in successive rows for Template:Math gives a triangular array called Pascal's triangle, satisfying the recurrence relation

(\binom{n}{k}) = (\binom{n - 1}{k - 1}) + (\binom{n - 1}{k}) .

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

The first form of the binomial coefficients can be generalized to $(\binom{z}{k})$ for any complex number Template:Mvar and integer Template:Math, and many of their properties continue to hold in this more general form.

History and notation

Andreas von Ettingshausen introduced the notation $(\binom{n}{k})$ in 1826,^[1] although the numbers were known centuries earlier (see Pascal's triangle). In about 1150, the Indian mathematician Bhaskaracharya gave an exposition of binomial coefficients in his book Līlāvatī.^[2]

Alternative notations include Template:Math, Template:Math, Template:Math, Template:Math,^[3] Template:Math, and Template:Math, in all of which the Template:Mvar stands for combinations or choices; the Template:Mvar notation means the number of ways to choose k out of n objects. Many calculators use variants of the Template:Nowrap because they can represent it on a single-line display. In this form the binomial coefficients are easily compared to the numbers of [[permutation#k-permutations of n|Template:Mvar-permutations of Template:Mvar]], written as Template:Math, etc.

Definition and interpretations

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

For natural numbers (taken to include 0) Template:Mvar and Template:Mvar, the binomial coefficient $(\binom{n}{k})$ can be defined as the coefficient of the monomial Template:Math in the expansion of Template:Math. The same coefficient also occurs (if Template:Math) in the binomial formula Template:NumBlk2 (valid for any elements Template:Mvar, Template:Mvar 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 Template:Mvar objects can be chosen from among Template:Mvar objects; more formally, the number of Template:Mvar-element subsets (or Template:Mvar-combinations) of an Template:Mvar-element set. This number can be seen as equal to the one of the first definition, independently of any of the formulas below to compute it: if in each of the Template:Mvar factors of the power Template:Math one temporarily labels the term Template:Mvar with an index Template:Mvar (running from Template:Math to Template:Mvar), then each subset of Template:Mvar indices gives after expansion a contribution Template:Math, and the coefficient of that monomial in the result will be the number of such subsets. This shows in particular that $(\binom{n}{k})$ is a natural number for any natural numbers Template:Mvar and Template:Mvar. There are many other combinatorial interpretations of binomial coefficients (counting problems for which the answer is given by a binomial coefficient expression), for instance the number of words formed of Template:Mvar bits (digits 0 or 1) whose sum is Template:Mvar is given by $(\binom{n}{k})$ , while the number of ways to write $k = a_{1} + a_{2} + \dots + a_{n}$ where every Template:Math is a nonnegative integer is given by Template:Tmath. Most of these interpretations can be shown to be equivalent to counting Template:Mvar-combinations.

Computing the value of binomial coefficients

Several methods exist to compute the value of $(\binom{n}{k})$ without actually expanding a binomial power or counting Template:Mvar-combinations.

Recursive formula

One method uses the recursive, purely additive formula $(\binom{n}{k}) = (\binom{n - 1}{k - 1}) + (\binom{n - 1}{k})$ for all integers $n, k$ such that $1 \leq k < n,$ with boundary values $(\binom{n}{0}) = (\binom{n}{n}) = 1$ for all integers Template:Math.

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

Multiplicative formula

A more efficient method to compute individual binomial coefficients is given by the formula $(\binom{n}{k}) = \frac{n^{\underline{k}}}{k!} = \frac{n (n - 1) (n - 2) \dots (n - (k - 1))}{k (k - 1) (k - 2) \dots 1} = \prod_{i = 1}^{k} \frac{n + 1 - i}{i},$ where the numerator of the first fraction, $n^{\underline{k}}$ , 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 Template:Mvar distinct objects, retaining the order of selection, from a set of Template:Mvar objects. The denominator counts the number of distinct sequences that define the same Template:Mvar-combination when order is disregarded. This formula can also be stated in a recursive form. Using the "C" notation from above, $C_{n, k} = C_{n, k - 1} \cdot (n - k + 1) / k$ , where $C_{n, 0} = 1$ . It is readily derived by evaluating $C_{n, k} / C_{n, k - 1}$ and can intuitively be understood as starting at the leftmost coefficient of the $n$ -th row of Pascal's triangle, whose value is always $1$ , and recursively computing the next coefficient to its right until the $k$ -th one is reached.

Due to the symmetry of the binomial coefficients with regard to Template:Mvar and Template:Math, calculation of the above product, as well as the recursive relation, may be optimised by setting its upper limit to the smaller of Template:Mvar and Template:Math.

Factorial formula

Finally, though computationally unsuitable, there is the compact form, often used in proofs and derivations, which makes repeated use of the familiar factorial function: $(\binom{n}{k}) = \frac{n!}{k! (n - k)!} for 0 \leq k \leq n,$ where Template:Math denotes the factorial of Template:Mvar. This formula follows from the multiplicative formula above by multiplying numerator and denominator by Template:Math; as a consequence it involves many factors common to numerator and denominator. It is less practical for explicit computation (in the case that Template:Mvar is small and Template:Mvar is large) unless common factors are first cancelled (in particular since factorial values grow very rapidly). The formula does exhibit a symmetry that is less evident from the multiplicative formula (though it is from the definitions) Template:NumBlk2 which leads to a more efficient multiplicative computational routine. Using the falling factorial notation, $(\binom{n}{k}) = {\begin{matrix} n^{\underline{k}} / k! & if k \leq \frac{n}{2} \\ n^{\underline{n - k}} / (n - k)! & if k > \frac{n}{2} \end{matrix} .$

Generalization and connection to the binomial series

Template:Main 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: $(\binom{α}{k}) = \frac{α^{\underline{k}}}{k!} = \frac{α (α - 1) (α - 2) \dots (α - k + 1)}{k (k - 1) (k - 2) \dots 1} for k \in ℕ and arbitrary α .$

With this definition one has a generalization of the binomial formula (with one of the variables set to 1), which justifies still calling the $(\binom{α}{k})$ binomial coefficients: Template:NumBlk2

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 $(1 + X)^{α} (1 + X)^{β} = (1 + X)^{α + β} and ((1 + X)^{α})^{β} = (1 + X)^{α β} .$

If α is a nonnegative integer n, then all terms with Template:Math are zero,^[5] 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

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

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

Template:Main

Pascal's rule is the important recurrence relation Template:NumBlk2 which can be used to prove by mathematical induction that $(\binom{n}{k})$ 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 Template:Mvar contains the numbers $(\binom{n}{k})$ for Template:Math. It is constructed by first placing 1s in the outermost positions, and then filling each inner position with the sum of the two numbers directly above. This method allows the quick calculation of binomial coefficients without the need for fractions or multiplications. For instance, by looking at row number 5 of the triangle, one can quickly read off that

(x + y)^{5} = \underline{1} x^{5} + \underline{5} x^{4} y + \underline{10} x^{3} y^{2} + \underline{10} x^{2} y^{3} + \underline{5} x y^{4} + \underline{1} y^{5} .

Combinatorics and statistics

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

There are $(\binom{n}{k})$ ways to choose k elements from a set of n elements. See Combination.
There are $(\binom{n + k - 1}{k})$ ways to choose k elements from a set of n elements if repetitions are allowed. See Multiset.
There are $(\binom{n + k}{k})$ strings containing k ones and n zeros.
There are $(\binom{n + 1}{k})$ strings consisting of k ones and n zeros such that no two ones are adjacent.^[6]
The Catalan numbers are $\frac{1}{n + 1} (\binom{2 n}{n}) .$
The binomial distribution in statistics is $(\binom{n}{k}) p^{k} (1 - p)^{n - k} .$

Binomial coefficients as polynomials

For any nonnegative integer k, the expression $(\binom{t}{k})$ can be written as a polynomial with denominator Template:Math:

(\binom{t}{k}) = \frac{t^{\underline{k}}}{k!} = \frac{t (t - 1) (t - 2) \dots (t - k + 1)}{k (k - 1) (k - 2) \dots 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 $(\binom{t}{k})$ can be characterized as the unique degree k polynomial Template:Math satisfying Template:Math and Template:Math.

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

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

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

\frac{d}{d t} (\binom{t}{k}) = (\binom{t}{k}) \sum_{i = 0}^{k - 1} \frac{1}{t - i} .

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

\frac{d}{d t} (\binom{t}{k}) = \sum_{i = 0}^{k - 1} \frac{(- 1)^{k - i - 1}}{k - i} (\binom{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, because the binomial coefficients consist of one polynomial of each degree. The coefficient a_k is the kth difference of the sequence p(0), p(1), ..., p(k). Explicitly,^[7] Template:NumBlk2

Integer-valued polynomials

Template:Main Each polynomial $(\binom{t}{k})$ is integer-valued: it has an integer value at all integer inputs $t$ . (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, (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 Template:Math can be rewritten as

9 (\binom{t}{2}) + 6 (\binom{t}{1}) + 0 (\binom{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:NumBlk2 and, with a little more work,

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

We can also get

(\binom{n - 1}{k}) = \frac{n - k}{n} (\binom{n}{k}) .

Moreover, the following may be useful:

(\binom{n}{k}) (\binom{k}{j}) = (\binom{n}{j}) (\binom{n - j}{k - j}) = (\binom{n}{k - j}) (\binom{n - k + j}{j}) .

For constant n, we have the following recurrence:

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

To sum up, we have

(\binom{n}{k}) = (\binom{n}{n - k}) = \frac{n - k + 1}{k} (\binom{n}{k - 1}) = \frac{n}{n - k} (\binom{n - 1}{k})

= \frac{n}{k} (\binom{n - 1}{k - 1}) = \frac{n}{n - 2 k} ((\binom{n - 1}{k}) - (\binom{n - 1}{k - 1})) = (\binom{n - 1}{k}) + (\binom{n - 1}{k - 1}) .

Sums of the binomial coefficients

The formula Template:NumBlk2 says that the elements in the Template:Mvarth row of Pascal's triangle always add up to 2 raised to the Template:Mvarth power. This is obtained from the binomial theorem (Template:EquationNote) by setting Template:Math and Template:Math. The formula also has a natural combinatorial interpretation: the left side sums the number of subsets of {1, ..., n} of sizes k = 0, 1, ..., n, giving the total number of subsets. (That is, the left side counts the power set of {1, ..., n}.) However, these subsets can also be generated by successively choosing or excluding each element 1, ..., n; the n independent binary choices (bit-strings) allow a total of $2^{n}$ choices. The left and right sides are two ways to count the same collection of subsets, so they are equal.

The formulas Template:NumBlk2 and

\sum_{k = 0}^{n} k^{2} (\binom{n}{k}) = (n + n^{2}) 2^{n - 2}

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

The Chu–Vandermonde identity, which holds for any complex values m and n and any non-negative integer k, is Template:NumBlk2 and can be found by examination of the coefficient of $x^{k}$ in the expansion of Template:Math using equation (Template:EquationNote). When Template:Math, equation (Template:EquationNote) reduces to equation (Template:EquationNote). In the special case Template:Math, using (Template:EquationNote), the expansion (Template:EquationNote) becomes (as seen in Pascal's triangle at right) Template:Image frame Template:NumBlk2 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 Template:Math, is Template:NumBlk2 The proof is similar, but uses the binomial series expansion (Template:EquationNote) with negative integer exponents. When Template:Math, equation (Template:EquationNote) gives the hockey-stick identity

\sum_{m = k}^{n} (\binom{m}{k}) = (\binom{n + 1}{k + 1})

and its relative

\sum_{r = 0}^{m} (\binom{n + r}{r}) = (\binom{n + m + 1}{m}) .

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

\sum_{k = 0}^{⌊ n / 2 ⌋} (\binom{n - k}{k}) = F (n + 1) .

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

Multisections of sums

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

(\binom{n}{t}) + (\binom{n}{t + s}) + (\binom{n}{t + 2 s}) + \dots = \frac{1}{s} \sum_{j = 0}^{s - 1} {(2 \cos \frac{π j}{s})}^{n} \cos \frac{π (n - 2 t) j}{s} .

For small Template:Mvar, these series have particularly nice forms; for example,^[8]

(\binom{n}{0}) + (\binom{n}{3}) + (\binom{n}{6}) + \dots = \frac{1}{3} (2^{n} + 2 \cos \frac{n π}{3})

(\binom{n}{1}) + (\binom{n}{4}) + (\binom{n}{7}) + \dots = \frac{1}{3} (2^{n} + 2 \cos \frac{(n - 2) π}{3})

(\binom{n}{2}) + (\binom{n}{5}) + (\binom{n}{8}) + \dots = \frac{1}{3} (2^{n} + 2 \cos \frac{(n - 4) π}{3})

(\binom{n}{0}) + (\binom{n}{4}) + (\binom{n}{8}) + \dots = \frac{1}{2} (2^{n - 1} + 2^{\frac{n}{2}} \cos \frac{n π}{4})

(\binom{n}{1}) + (\binom{n}{5}) + (\binom{n}{9}) + \dots = \frac{1}{2} (2^{n - 1} + 2^{\frac{n}{2}} \sin \frac{n π}{4})

(\binom{n}{2}) + (\binom{n}{6}) + (\binom{n}{10}) + \dots = \frac{1}{2} (2^{n - 1} - 2^{\frac{n}{2}} \cos \frac{n π}{4})

(\binom{n}{3}) + (\binom{n}{7}) + (\binom{n}{11}) + \dots = \frac{1}{2} (2^{n - 1} - 2^{\frac{n}{2}} \sin \frac{n π}{4})

Partial sums

Although there is no closed formula for partial sums

\sum_{j = 0}^{k} (\binom{n}{j})

of binomial coefficients,^[9] one can again use (Template:EquationNote) and induction to show that for Template:Math,

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

with special case^[10]

\sum_{j = 0}^{n} (- 1)^{j} (\binom{n}{j}) = 0

for Template:Math. This latter result is also a special case of the result from the theory of finite differences that for any polynomial P(x) of degree less than n,^[11]

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

Differentiating (Template:EquationNote) k times and setting x = −1 yields this for $P (x) = x (x - 1) \dots (x - k + 1)$ , 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, Template:NumBlk2 where $a_{n}$ is the coefficient of degree n in P(x).

More generally for (Template:EquationNote),

\sum_{j = 0}^{n} (- 1)^{j} (\binom{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 Template:Tmath instead of Template:Tmath, and observing that Template:Tmath still has degree less than or equal to n, and that its coefficient of degree n is dⁿa_n.

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.

Identities with combinatorial proofs

Many identities involving binomial coefficients can be proved by combinatorial means. For example, for nonnegative integers $n \geq q$ , the identity

\sum_{k = q}^{n} (\binom{n}{k}) (\binom{k}{q}) = 2^{n - q} (\binom{n}{q})

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

In Pascal's identity

(\binom{n}{k}) = (\binom{n - 1}{k - 1}) + (\binom{n - 1}{k}),

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

\sum_{k = 0}^{n} {(\binom{n}{k})}^{2} = (\binom{2 n}{n}) .

Suppose you have $2 n$ empty squares arranged in a row and you want to mark (select) n of them. There are $(\binom{2 n}{n})$ 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} (\binom{n}{k}) (\binom{n}{n - k}) = (\binom{2 n}{n}) .

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

If one denotes by Template:Math the sequence of Fibonacci numbers, indexed so that Template:Math, then the identity $\sum_{k = 0}^{⌊ \frac{n}{2} ⌋} (\binom{n - k}{k}) = F (n)$ has the following combinatorial proof.^[12] One may show by induction that Template:Math counts the number of ways that a Template:Math strip of squares may be covered by Template:Math and Template:Math tiles. On the other hand, if such a tiling uses exactly Template:Mvar of the Template:Math tiles, then it uses Template:Math of the Template:Math tiles, and so uses Template:Math tiles total. There are $(\binom{n - k}{k})$ ways to order these tiles, and so summing this coefficient over all possible values of Template:Mvar gives the identity.

Sum of coefficients row

Template:See also

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

Dixon's identity

Dixon's identity is

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

or, more generally,

\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!},

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

Continuous identities

Certain trigonometric integrals have values expressible in terms of binomial coefficients: For any $m, n \in ℕ,$

\int_{- π}^{π} \cos ((2 m - n) x) \cos^{n} (x) d x = \frac{π}{2^{n - 1}} (\binom{n}{m})

\int_{- π}^{π} \sin ((2 m - n) x) \sin^{n} (x) d x = {\begin{matrix} (- 1)^{m + (n + 1) / 2} \frac{π}{2^{n - 1}} (\binom{n}{m}), & n odd \\ 0, & otherwise \end{matrix}

\int_{- π}^{π} \cos ((2 m - n) x) \sin^{n} (x) d x = {\begin{matrix} (- 1)^{m + (n / 2)} \frac{π}{2^{n - 1}} (\binom{n}{m}), & n even \\ 0, & otherwise \end{matrix}

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 $(\binom{n - 1}{k}) \equiv (- 1)^{k} mod n$ for every k with $0 \leq k \leq n - 1 .$ 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

(\binom{n - 1}{k}) = \frac{(n - 1) (n - 2) \dots (n - k)}{1 \cdot 2 \dots k} = \prod_{i = 1}^{k} \frac{n - i}{i} \equiv \prod_{i = 1}^{k} \frac{- i}{i} = (- 1)^{k} mod n .

Generating functions

Ordinary generating functions

For a fixed Template:Mvar, the ordinary generating function of the sequence $(\binom{n}{0}), (\binom{n}{1}), (\binom{n}{2}), \dots$ is

\sum_{k = 0}^{\infty} (\binom{n}{k}) x^{k} = (1 + x)^{n} .

For a fixed Template:Mvar, the ordinary generating function of the sequence $(\binom{0}{k}), (\binom{1}{k}), (\binom{2}{k}), \dots,$ is

\sum_{n = 0}^{\infty} (\binom{n}{k}) y^{n} = \frac{y^{k}}{(1 - y)^{k + 1}} .

The bivariate generating function of the binomial coefficients is

\sum_{n = 0}^{\infty} \sum_{k = 0}^{n} (\binom{n}{k}) x^{k} y^{n} = \frac{1}{1 - y - x y} .

A symmetric bivariate generating function of the binomial coefficients is

\sum_{n = 0}^{\infty} \sum_{k = 0}^{\infty} (\binom{n + k}{k}) x^{k} y^{n} = \frac{1}{1 - x - y} .

which is the same as the previous generating function after the substitution $x \to x y$ .

Exponential generating function

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

\sum_{n = 0}^{\infty} \sum_{k = 0}^{\infty} (\binom{n + k}{k}) \frac{x^{k} y^{n}}{(n + k)!} = e^{x + y} .

Divisibility properties

Template: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 $(\binom{m + n}{m})$ 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 $(\binom{n}{k})$ equals the number of nonnegative integers j such that the fractional part of k/p^j is greater than the fractional part of n/p^j. It can be deduced from this that $(\binom{n}{k})$ is divisible by n/gcd(n,k). In particular therefore it follows that p divides $(\binom{p^{r}}{s})$ for all positive integers r and s such that Template:Math. However this is not true of higher powers of p: for example 9 does not divide $(\binom{9}{6})$ .

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 $(\binom{n}{k})$ with n < N such that d divides $(\binom{n}{k})$ . Then

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

Since the number of binomial coefficients $(\binom{n}{k})$ 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:^[13]

(\binom{n + k}{k})

divides

\frac{lcm (n, n + 1, \dots, n + k)}{n}

.

(\binom{n + k}{k})

is a multiple of

\frac{lcm (n, n + 1, \dots, n + k)}{n \cdot lcm ((\binom{k}{0}), (\binom{k}{1}), \dots, (\binom{k}{k}))}

.

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

(\binom{n}{1}), (\binom{n}{2}), \dots, (\binom{n}{n - 1})

are divisible by n.

Proof: When p is prime, p divides

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

for all Template:Math

because $(\binom{p}{k})$ 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 Template:Math. Then Template:Math and

(\binom{n}{p}) = \frac{n (n - 1) (n - 2) \dots (n - p + 1)}{p!} = \frac{k (n - 1) (n - 2) \dots (n - p + 1)}{(p - 1)!} \equiv̸ 0 (\mod n)

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

Bounds and asymptotic formulas

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

From the divisibility properties we can infer that $\frac{lcm (n - k, \dots, n)}{(n - k) \cdot lcm ((\binom{k}{0}), \dots, (\binom{k}{k}))} \leq (\binom{n}{k}) \leq \frac{lcm (n - k, \dots, n)}{n - k},$ where both equalities can be achieved.^[13]

The following bounds are useful in information theory:^[14]Template:Rp $\frac{1}{n + 1} 2^{n H (k / n)} \leq (\binom{n}{k}) \leq 2^{n H (k / n)}$ where $H (p) = - p \log_{2} (p) - (1 - p) \log_{2} (1 - p)$ is the binary entropy function. It can be further tightened to $\sqrt{\frac{n}{8 k (n - k)}} 2^{n H (k / n)} \leq (\binom{n}{k}) \leq \sqrt{\frac{n}{2 π k (n - k)}} 2^{n H (k / n)}$ for all $1 \leq k \leq n - 1$ .^[15]Template:Rp

Both n and k large

Stirling's approximation yields the following approximation, valid when $n - k, k$ both tend to infinity: $(\binom{n}{k}) \sim \sqrt{\frac{n}{2 π k (n - k)}} \cdot \frac{n^{n}}{k^{k} (n - k)^{n - k}}$ 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 $n$ is sufficiently large, one has $(\binom{2 n}{n}) \sim \frac{2^{2 n}}{\sqrt{n π}}$ and $\sqrt{n} (\binom{2 n}{n}) \geq 2^{2 n - 1}$ . More generally, for Template:Math and Template:Math (again, by applying Stirling's formula to the factorials in the binomial coefficient), $\sqrt{n} (\binom{m n}{n}) \geq \frac{m^{m (n - 1) + 1}}{(m - 1)^{(m - 1) (n - 1)}} .$

If n is large and k is linear in n, various precise asymptotic estimates exist for the binomial coefficient $(\binom{n}{k})$ . For example, if $| n / 2 - k | = o (n^{2 / 3})$ then $(\binom{n}{k}) \sim (\binom{n}{\frac{n}{2}}) e^{- d^{2} / (2 n)} \sim \frac{2^{n}}{\sqrt{\frac{1}{2} n π}} e^{- d^{2} / (2 n)}$ where d = n − 2k.^[16]

Template:Mvar much larger than Template:Mvar

If Template:Mvar is large and Template:Mvar is Template:Math (that is, if Template:Math), then $(\binom{n}{k}) \sim {(\frac{n e}{k})}^{k} \cdot (2 π k)^{- 1 / 2} \cdot \exp (- \frac{k^{2}}{2 n} (1 + o (1)))$ where again Template:Mvar is the little o notation.^[17]

Sums of binomial coefficients

A simple and rough upper bound for the sum of binomial coefficients can be obtained using the binomial theorem: $\sum_{i = 0}^{k} (\binom{n}{i}) \leq \sum_{i = 0}^{k} n^{i} \cdot 1^{k - i} \leq (1 + n)^{k}$ More precise bounds are given by $\frac{1}{\sqrt{8 n ε (1 - ε)}} \cdot 2^{H (ε) \cdot n} \leq \sum_{i = 0}^{k} (\binom{n}{i}) \leq 2^{H (ε) \cdot n},$ valid for all integers $n > k \geq 1$ with $ε ≐ k / n \leq 1 / 2$ .^[18]

Generalized binomial coefficients

The infinite product formula for the gamma function also gives an expression for binomial coefficients $(- 1)^{k} (\binom{z}{k}) = (\binom{- z + k - 1}{k}) = \frac{1}{Γ (- z)} \frac{1}{(k + 1)^{z + 1}} \prod_{j = k + 1} \frac{{(1 + \frac{1}{j})}^{- z - 1}}{1 - \frac{z + 1}{j}}$ which yields the asymptotic formulas $(\binom{z}{k}) \approx \frac{(- 1)^{k}}{Γ (- z) k^{z + 1}} and (\binom{z + k}{k}) = \frac{k^{z}}{Γ (z + 1)} (1 + \frac{z (z + 1)}{2 k} + 𝒪 (k^{- 2}))$ as $k \to \infty$ .

This asymptotic behaviour is contained in the approximation $(\binom{z + k}{k}) \approx \frac{e^{z (H_{k} - γ)}}{Γ (z + 1)}$ as well. (Here $H_{k}$ is the k-th harmonic number and $γ$ is the Euler–Mascheroni constant.)

Further, the asymptotic formula $\frac{(\binom{z + k}{j})}{(\binom{k}{j})} \to {(1 - \frac{j}{k})}^{- z} and \frac{(\binom{j}{j - k})}{(\binom{j - z}{j - k})} \to {(\frac{j}{k})}^{z}$ hold true, whenever $k \to \infty$ and $j / k \to x$ for some complex number $x$ .

Generalizations

Generalization to multinomials

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

(\binom{n}{k_{1}, k_{2}, \dots, k_{r}}) = \frac{n!}{k_{1}! k_{2}! \dots k_{r}!}

where

\sum_{i = 1}^{r} k_{i} = n .

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

(x_{1} + x_{2} + \dots + x_{r})^{n} .

The case r = 2 gives binomial coefficients:

(\binom{n}{k_{1}, k_{2}}) = (\binom{n}{k_{1}, n - k_{1}}) = (\binom{n}{k_{1}}) = (\binom{n}{k_{2}}) .

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

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

(\binom{n}{k_{1}, k_{2}, \dots, k_{r}}) = (\binom{n - 1}{k_{1} - 1, k_{2}, \dots, k_{r}}) + (\binom{n - 1}{k_{1}, k_{2} - 1, \dots, k_{r}}) + \dots + (\binom{n - 1}{k_{1}, k_{2}, \dots, k_{r} - 1})

and symmetry:

(\binom{n}{k_{1}, k_{2}, \dots, k_{r}}) = (\binom{n}{k_{σ_{1}}, k_{σ_{2}}, \dots, k_{σ_{r}}})

where $(σ_{i})$ 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{matrix} (\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{matrix}

Binomial coefficient with Template:Math

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 $k$ :

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

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

\sqrt{1 + x} = \sum_{k \geq 0} (\binom{1 / 2}{k}) x^{k} .

Products of binomial coefficients

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

(\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}),

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

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

\prod_{k = 0}^{n} (\binom{n}{k}) = \prod_{k = 1}^{n} k^{2 k - n - 1} .

Partial fraction decomposition

The partial fraction decomposition of the reciprocal is given by

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

Newton's binomial series

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

(1 + z)^{α} = \sum_{n = 0}^{\infty} (\binom{α}{n}) z^{n} = 1 + (\binom{α}{1}) z + (\binom{α}{2}) z^{2} + \dots .

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

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

\frac{1}{(1 - z)^{α + 1}} = \sum_{n = 0}^{\infty} (\binom{n + α}{n}) z^{n}

where the identity

(\binom{n}{k}) = (- 1)^{k} (\binom{k - n - 1}{k})

is applied.

Multiset (rising) binomial coefficient

Template:Main Binomial coefficients count subsets of prescribed size from a given set. A related combinatorial problem is to count multisets of prescribed size with elements drawn from a given set, that is, to count the number of ways to select a certain number of elements from a given set with the possibility of selecting the same element repeatedly. The resulting numbers are called multiset coefficients;^[19] the number of ways to "multichoose" (i.e., choose with replacement) k items from an n element set is denoted $((\binom{n}{k}))$ .

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

Multiset coefficients may be expressed in terms of binomial coefficients by the rule $(\binom{f}{k}) = ((\binom{r}{k})) = (\binom{r + k - 1}{k}) .$ One possible alternative characterization of this identity is as follows: We may define the falling factorial as $(f)_{k} = f^{\underline{k}} = (f - k + 1) \dots (f - 3) \cdot (f - 2) \cdot (f - 1) \cdot f,$ and the corresponding rising factorial as $r^{(k)} = r^{\overline{k}} = r \cdot (r + 1) \cdot (r + 2) \cdot (r + 3) \dots (r + k - 1);$ so, for example, $17 \cdot 18 \cdot 19 \cdot 20 \cdot 21 = (21)_{5} = 2 1^{\underline{5}} = 1 7^{\overline{5}} = 1 7^{(5)} .$ Then the binomial coefficients may be written as $(\binom{f}{k}) = \frac{(f)_{k}}{k!} = \frac{(f - k + 1) \dots (f - 2) \cdot (f - 1) \cdot f}{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \dots k},$ while the corresponding multiset coefficient is defined by replacing the falling with the rising factorial: $((\binom{r}{k})) = \frac{r^{(k)}}{k!} = \frac{r \cdot (r + 1) \cdot (r + 2) \dots (r + k - 1)}{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \dots k} .$

Generalization to negative integers n

Template:Pascal triangle extended.svg For any n,

\begin{matrix} (\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) \dots (n + k - 1)}{k!} \\ = (- 1)^{k} (\binom{n + k - 1}{k}) \\ = (- 1)^{k} ((\binom{n}{k})) . \end{matrix}

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

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

\begin{matrix} (\binom{- 4}{7}) & = \frac{- 10 \cdot - 9 \cdot - 8 \cdot - 7 \cdot - 6 \cdot - 5 \cdot - 4}{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7} \\ = (- 1)^{7} \frac{4 \cdot 5 \cdot 6 \cdot 7 \cdot 8 \cdot 9 \cdot 10}{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7} \\ = ((\binom{- 7}{7})) ((\binom{4}{7})) = (\binom{- 1}{7}) (\binom{10}{7}) . \end{matrix}

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

(\binom{x}{y}) = \frac{Γ (x + 1)}{Γ (y + 1) Γ (x - y + 1)} = \frac{1}{(x + 1) B (y + 1, x - y + 1)} .

This definition inherits these following additional properties from $Γ$ :

(\binom{x}{y}) = \frac{\sin (y π)}{\sin (x π)} (\binom{- y - 1}{- x - 1}) = \frac{\sin ((x - y) π)}{\sin (x π)} (\binom{y - x - 1}{y});

moreover,

(\binom{x}{y}) \cdot (\binom{y}{x}) = \frac{\sin ((x - y) π)}{(x - y) π} .

The resulting function has been little-studied, apparently first being graphed in Template:Harv. Notably, many binomial identities fail: $(\binom{n}{m}) = (\binom{n}{n - m})$ but $(\binom{- n}{m}) \neq (\binom{- n}{- n - m})$ 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:

(\binom{α}{β}) = | {B \subseteq A : | B | = β} |

where Template:Mvar is some set with cardinality $α$ . 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 $α$ , $(\binom{α}{β})$ 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 $(\binom{α}{α}) = 2^{α}$ for any infinite cardinal $α$ .

Notes

Template:Reflist

References

Template:Refbegin

Template:Refend

External links

Template:PlanetMath attribution

↑ Template:Harvtxt
↑ Lilavati Section 6, Chapter 4 (see Template:Harvtxt).
↑ Template:Harvnb
↑ See Template:Harv, which also defines $(\binom{n}{k}) = 0$ for $k < 0$ . Alternative generalizations, such as to two real or complex valued arguments using the Gamma function assign nonzero values to $(\binom{n}{k})$ for $k < 0$ , 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).
↑ When $α = n$ is a nonnegative integer, $(\binom{n}{k}) = 0$ for $k > n$ because the $(k = n + 1)$ -th factor of the numerator is $n - (n + 1) + 1 = 0$ . Thus, the $k$ -th term is a zero product for all $k \geq n + 1$ .
↑ Template:Cite journal
↑ 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.
↑ Template:Harvtxt.
↑ Template:Citation.
↑ see induction developed in eq (7) p. 1389 in Template:Citation.
↑ Template:Cite journal
↑ Template:Harvnb
↑ ^13.0 ^13.1 Template:Cite journal
↑ Template:Cite book
↑ Template:Cite book
↑ Template:Cite book
↑ Template:Cite book
↑ see e.g. Template:Harvtxt or Template:Harvtxt.
↑ Template:Citation.

[1] Template:Harvtxt

[2] Lilavati Section 6, Chapter 4 (see Template:Harvtxt).

[3] Template:Harvnb

[4] See Template:Harv, which also defines $(\binom{n}{k}) = 0$ for $k < 0$ . Alternative generalizations, such as to two real or complex valued arguments using the Gamma function assign nonzero values to $(\binom{n}{k})$ for $k < 0$ , 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).

[5] When $α = n$ is a nonnegative integer, $(\binom{n}{k}) = 0$ for $k > n$ because the $(k = n + 1)$ -th factor of the numerator is $n - (n + 1) + 1 = 0$ . Thus, the $k$ -th term is a zero product for all $k \geq n + 1$ .

[6] Template:Cite journal

[7] 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.

[8] Template:Harvtxt.

[9] Template:Citation.

[10] see induction developed in eq (7) p. 1389 in Template:Citation.

[11] Template:Cite journal

[12] Template:Harvnb

[Farhi2007-13] 13.0 ^13.1 Template:Cite journal

[cover2006-14] Template:Cite book

[macwilliams1977-15] Template:Cite book

[16] Template:Cite book

[17] Template:Cite book

[18] see e.g. Template:Harvtxt or Template:Harvtxt.

[19] Template:Citation.

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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

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