Binomial coefficient

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 \(\tbinom{n}{k}\) or ⁠\(C(n, k)\)⁠. It is the coefficient of the x term in the polynomial expansion of the binomial power (1 + x); this coefficient can be computed by the multiplicative formula

\[\binom nk = \frac{n\times(n-1)\times\cdots\times(n-k+1)}{k\times(k-1)\times\cdots\times1},\]

which using factorial notation can be compactly expressed as

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

For example, the fourth power of 1 + x is \[\begin{align} (1 + x)^4 &= \tbinom{4}{0} x^0 + \tbinom{4}{1} x^1 + \tbinom{4}{2} x^2 + \tbinom{4}{3} x^3 + \tbinom{4}{4} x^4 \\ &= 1 + 4x + 6 x^2 + 4x^3 + x^4, \end{align}\] and the binomial coefficient \(\tbinom{4}{2} =\tfrac{4\times 3}{2\times1} = \tfrac{4!}{2!2!} = 6\) is the coefficient of the x 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 \[\binom{n}{k} = \binom{n-1}{k-1} + \binom{n-1}{k} .\]

The binomial coefficients occur in many areas of mathematics, and especially in combinatorics. In combinatorics the symbol \(\tbinom{n}{k}\) 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}.

Condensed: the full section is in Wikipedia.

Andreas von Ettingshausen introduced the notation \(\tbinom nk\) in 1826, although the numbers were known centuries earlier (see Pascal's triangle). Around 1150, the Indian mathematician Bhaskaracharya gave an exposition of binomial coefficients in his book Līlāvatī.

Alternative notations include C(n, k), _nC_k, C_k, C
_n, C
_k, and C_n,k, in all of which the C stands for combinations or choices; the C notation means the number of ways to choose k out of n objects. 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 the numbers of k-permutations of n, written as P(n, k), etc.

For natural numbers (taken to include 0) n and k, the binomial coefficient \(\tbinom nk\) can be defined as the coefficient of the monomial X in the expansion of (1 + X). 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 that 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) 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, 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 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.

One method uses the recursive, purely additive formula \[\binom nk = \binom{n-1}{k-1} + \binom{n-1}k\] for all integers \(n,k\) such that ⁠\(1 \le k < n\)⁠, with boundary values \[\binom n0 = \binom nn = 1\] 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 in (1 + X)(1 + X). As there is zero X or X in (1 + X), 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.

A more efficient method to compute individual binomial coefficients is given by the formula \[\binom nk = \frac{n^{\underline{k}}}{k!} = \frac{n(n-1)(n-2)\cdots(n-(k-1))}{k(k-1)(k-2)\cdots 1} = \prod_{i=1}^k\frac{ n+1-i}{ i},\] 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 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. 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 k and n − k, calculation of the above product, as well as the recursive relation, may be optimised by setting its upper limit to the smaller of k and n − k.

Finally, 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 (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)

which leads to a more efficient multiplicative computational routine. Using the falling factorial notation, \[\binom nk = \begin{cases} n^{\underline{k}}/k! & \text{if }\ k \le \frac{n}{2} \\ n^{\underline{n-k}}/(n-k)! & \text{if }\ k > \frac{n}{2} \end{cases}.\]

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

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

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 rule is the important recurrence relation

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.

Pascal's rule also gives rise to Pascal's triangle:

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

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.
• The Catalan numbers are ⁠\(\tfrac{1}{n+1}\tbinom{2n}{n}\)⁠.
• The binomial distribution in statistics is ⁠\(\tbinom n k p^k (1-p)^{n-k}\)⁠.

For any nonnegative integer k, the expression \(\binom{t}{k}\) can be written as a polynomial with denominator k!: \[\binom{t}{k} = \frac{t^\underline{k}}{k!} = \frac{t(t-1)(t-2)\cdots(t-k+1)}{k(k-1)(k-2)\cdots2 \cdot 1};\] 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: \[\binom{t}{k} = \sum_{i=0}^k s(k,i)\frac{t^i}{k!}.\] The derivative of \(\tbinom{t}{k}\) can be calculated by logarithmic differentiation: \[\frac{\mathrm{d}}{\mathrm{d}t} \binom{t}{k} = \binom{t}{k} \sum_{i=0}^{k-1} \frac{1}{t-i}.\] This can cause a problem when evaluated at integers from \(0\) to ⁠\(t-1\)⁠, but using identities below we can compute the derivative as: \[\frac{\mathrm{d}}{\mathrm{d}t} \binom{t}{k} = \sum_{i=0}^{k-1} \frac{(-1)^{k-i-1}}{k-i} \binom{t}{i}.\]

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,

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.

The factorial formula facilitates relating nearby binomial coefficients. For instance, if k is a positive integer and n is arbitrary, then

and, with a little more work, \[\binom {n-1}{k} - \binom{n-1}{k-1} = \frac{n-2k}{n} \binom{n}{k}.\]

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

The formula

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.

The formulas

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 x (twice for the latter) and then substituting x = y = 1.

The Chu-Vandermonde identity, which holds for any complex values m and n and any non-negative integer k, is

and can be found by examination of the coefficient of \(x^k\) in the expansion of (1 + x)(1 + x) = (1 + x) 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)

where the term on the right side is a central binomial coefficient.

Condensed: the full section is in Wikipedia.