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 n ≥ k ≥ 0 and is written ${\tbinom {n}{k}}.$ It is the coefficient of the x^{k} term in the polynomial expansion of the binomialpower(1 + x)^{n}; this coefficient can be computed by the multiplicative formula
and the binomial coefficient ${\tbinom {4}{2}}={\tfrac {4\times 3}{2\times 1}}={\tfrac {4!}{2!2!}}=6$ is the coefficient of the x^{2} term.
Arranging the numbers ${\tbinom {n}{0}},{\tbinom {n}{1}},\ldots ,{\tbinom {n}{n}}$ in successive rows for n = 0, 1, 2, ... gives a triangular array called Pascal's triangle, satisfying the recurrence relation
The binomial coefficients occur in many areas of mathematics, and especially in combinatorics. The symbol ${\tbinom {n}{k}}$ is usually read as "n choose k" because there are ${\tbinom {n}{k}}$ ways to choose an (unordered) subset of k elements from a fixed set of n elements. For example, there are ${\tbinom {4}{2}}=6$ ways to choose 2 elements from {1, 2, 3, 4}, namely {1, 2}, {1, 3}, {1, 4}, {2, 3}, {2, 4} and {3, 4}.
The binomial coefficients can be generalized to ${\tbinom {z}{k}}$ for any complex numberz and integer k ≥ 0, and many of their properties continue to hold in this more general form.
History and notationedit
Andreas von Ettingshausen introduced the notation ${\tbinom {n}{k}}$ 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 C(n, k), _{n}C_{k}, ^{n}C_{k}, C^{k} _{n},^{[3]}C^{n} _{k}, and C_{n,k}, in all of which the C stands for combinations or choices. Many calculators use variants of the C notation because they can represent it on a single-line display. In this form the binomial coefficients are easily compared to k-permutations of n, written as P(n, k), etc.
Definition and interpretationsedit
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The first few binomial coefficients on a left-aligned Pascal's triangle
For natural numbers (taken to include 0) n and k, the binomial coefficient ${\tbinom {n}{k}}$ can be defined as the coefficient of the monomialX^{k} in the expansion of (1 + X)^{n}. The same coefficient also occurs (if k ≤ n) in the binomial formula
(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 X^{k}, and the coefficient of that monomial in the result will be the number of such subsets. This shows in particular that ${\tbinom {n}{k}}$ 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 nbits (digits 0 or 1) whose sum is k is given by ${\tbinom {n}{k}}$, while the number of ways to write $k=a_{1}+a_{2}+\cdots +a_{n}$ where every a_{i} is a nonnegative integer is given by ${\tbinom {n+k-1}{n-1}}$. Most of these interpretations can be shown to be equivalent to counting k-combinations.
Computing the value of binomial coefficientsedit
Several methods exist to compute the value of ${\tbinom {n}{k}}$ without actually expanding a binomial power or counting k-combinations.
Recursive formulaedit
One method uses the recursive, purely additive formula
for all integers $n,k$ such that $1\leq k<n,$
with boundary values
${\binom {n}{0}}={\binom {n}{n}}=1$
for all integers n ≥ 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 X^{k} in (1 + X)^{n−1}(1 + X). As there is zero X^{n+1} or X^{−1} in (1 + X)^{n}, one might extend the definition beyond the above boundaries to include ${\tbinom {n}{k}}=0$ 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 formulaedit
A more efficient method to compute individual binomial coefficients is given by the formula
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.
Due to the symmetry of the binomial coefficient with regard to k and n − k, calculation may be optimised by setting the upper limit of the product above to the smaller of k and n − k.
Factorial formulaedit
Finally, though computationally unsuitable, there is the compact form, often used in proofs and derivations, which makes repeated use of the familiar factorial function:
where n! denotes the factorial of n. This formula follows from the multiplicative formula above by multiplying numerator and denominator by (n − k)!; as a consequence it involves many factors common to numerator and denominator. It is less practical for explicit computation (in the case that k is small and n is large) unless common factors are first cancelled (in particular since factorial values grow very rapidly). The formula does exhibit a symmetry that is less evident from the multiplicative formula (though it is from the definitions)
Generalization and connection to the binomial seriesedit
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:
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:
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
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.
which can be used to prove by mathematical induction that ${\tbinom {n}{k}}$ 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.
Row number n contains the numbers ${\tbinom {n}{k}}$ for k = 0, …, n. It is constructed by first placing 1s in the outermost positions, and then filling each inner position with the sum of the two numbers directly above. This method allows the quick calculation of binomial coefficients without the need for fractions or multiplications. For instance, by looking at row number 5 of the triangle, one can quickly read off that
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.
Binomial coefficients as a basis for the space of polynomialsedit
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 ${\textstyle \sum _{k=0}^{d}a_{k}{\binom {t}{k}}}$ of binomial coefficients. The coefficient a_{k} is the kth difference of the sequence p(0), p(1), ..., p(k). Explicitly,^{[6]}
Each polynomial ${\tbinom {t}{k}}$ 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, (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.
Exampleedit
The integer-valued polynomial 3t(3t + 1) / 2 can be rewritten as
says that the elements in the nth row of Pascal's triangle always add up to 2 raised to the nth power. This is obtained from the binomial theorem (∗) by setting x = 1 and y = 1. The formula also has a natural combinatorial interpretation: the left side sums the number of subsets of {1, ..., n} of sizes k = 0, 1, ..., n, giving the total number of subsets. (That is, the left side counts the power set of {1, ..., n}.) However, these subsets can also be generated by successively choosing or excluding each element 1, ..., n; the n independent binary choices (bit-strings) allow a total of $2^{n}$ choices. The left and right sides are two ways to count the same collection of subsets, so they are equal.
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). In the special case n = 2m, k = m, using (1), the expansion (7) becomes (as seen in Pascal's triangle at right)
The proof is similar, but uses the binomial series expansion (2) with negative integer exponents.
When j = k, equation (9) gives the hockey-stick identity
for n > 0. This latter result is also a special case of the result from the theory of finite differences that for any polynomial P(x) of degree less than n,^{[10]}
$\sum _{j=0}^{n}(-1)^{j}{\binom {n}{j}}P(j)=0.$
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.
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 $Q(x)$ still has degree less than or equal to n, and that its coefficient of degree n is d^{n}a_{n}.
The series${\textstyle {\frac {k-1}{k}}\sum _{j=0}^{\infty }{\frac {1}{\binom {j+x}{k}}}={\frac {1}{\binom {x-1}{k-1}}}}$ is convergent for k ≥ 2. This formula is used in the analysis of the German tank problem. It follows from ${\textstyle {\frac {k-1}{k}}\sum _{j=0}^{M}{\frac {1}{\binom {j+x}{k}}}={\frac {1}{\binom {x-1}{k-1}}}-{\frac {1}{\binom {M+x}{k-1}}}}$ which is proved by induction on M.
Identities with combinatorial proofsedit
Many identities involving binomial coefficients can be proved by combinatorial means. For example, for nonnegative integers ${n}\geq {q}$, the identity
(which reduces to (6) when q = 1) can be given a double counting proof, as follows. The left side counts the number of ways of selecting a subset of [n] = {1, 2, ..., n} with at least q elements, and marking q elements among those selected. The right side counts the same thing, because there are ${\tbinom {n}{q}}$ 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.
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 (8) also has a combinatorial proof. The identity reads
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
has the following combinatorial proof.^{[11]} One may show by induction that F(n) counts the number of ways that a n × 1 strip of squares may be covered by 2 × 1 and 1 × 1 tiles. On the other hand, if such a tiling uses exactly k of the 2 × 1 tiles, then it uses n − 2k of the 1 × 1 tiles, and so uses n − k tiles total. There are ${\tbinom {n-k}{k}}$ ways to order these tiles, and so summing this coefficient over all possible values of k gives the identity.
Sum of coefficients rowedit
The number of k-combinations for all k, ${\textstyle \sum _{0\leq {k}\leq {n}}{\binom {n}{k}}=2^{n}}$, 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.
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.
Congruencesedit
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.
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 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 ${\tbinom {n}{k}}$
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 ${\tbinom {n}{k}}$ 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 < p^{r}. 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.
Binomial coefficients have divisibility properties related to least common multiples of consecutive integers. For example:^{[12]}
${\binom {p}{k}}={\frac {p\cdot (p-1)\cdots (p-k+1)}{k\cdot (k-1)\cdots 1}}$ for all 0 < k < p
because ${\tbinom {p}{k}}$ 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 k = n/p. Then 0 < p < n and
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 formulasedit
The following bounds for ${\tbinom {n}{k}}$ hold for all values of n and k such that 1 ≤ k ≤ n:
and each of these $k$ terms in this product is ${\textstyle \geq {\frac {n}{k}}}$. A similar argument can be made to show the second inequality. The final strict inequality is equivalent to ${\textstyle e^{k}>k^{k}/k!}$, that is clear since the RHS is a term of the exponential series ${\textstyle e^{k}=\sum _{j=0}^{\infty }k^{j}/j!}$.
From the divisibility properties we can infer that
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
If n is large and k is linear in n, various precise asymptotic estimates exist for the binomial coefficient ${\textstyle {\binom {n}{k}}}$. For example, if $|n/2-k|=o(n^{2/3})$ then
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:
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.
The product of all binomial coefficients in the nth row of the Pascal triangle is given by the formula:
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;^{[18]} the number of ways to "multichoose" (i.e., choose with replacement) k items from an n element set is denoted ${\textstyle \left(\!\!{\binom {n}{k}}\!\!\right)}$.
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
In particular, binomial coefficients evaluated at negative integers n are given by signed multiset coefficients. In the special case $n=-1$, this reduces to $(-1)^{k}={\binom {-1}{k}}=\left(\!\!{\binom {-k}{k}}\!\!\right).$
For example, if n = −4 and k = 7, then r = 4 and f = 10:
The resulting function has been little-studied, apparently first being graphed in (Fowler 1996). Notably, many binomial identities fail: ${\textstyle {\binom {n}{m}}={\binom {n}{n-m}}}$ but ${\textstyle {\binom {-n}{m}}\neq {\binom {-n}{-n-m}}}$ 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.
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$\alpha$, ${\textstyle {\alpha \choose \beta }}$ 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 ${\textstyle {\alpha \choose \alpha }=2^{\alpha }}$ for any infinite cardinal $\alpha$.
^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 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).
^Muir, Thomas (1902). "Note on Selected Combinations". Proceedings of the Royal Society of Edinburgh.
^Boardman, Michael (2004), "The Egg-Drop Numbers", Mathematics Magazine, 77 (5): 368–372, doi:10.2307/3219201, 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.
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This article incorporates material from the following PlanetMath articles, which are licensed under the Creative Commons Attribution/Share-Alike License: Binomial Coefficient, Upper and lower bounds to binomial coefficient, Binomial coefficient is an integer, Generalized binomial coefficients.