Binomial Coefficient

Commonly, a binomial coefficient is indexed by a pair of integers n ≥ k ≥ 0 and is written ${\displaystyle {\tbinom {n}{k}}.}$ It is the coefficient of the x^k term in the polynomial expansion of the binomial power (1 + x)ⁿ; this coefficient can be computed by the multiplicative formula

${\displaystyle {\binom {n}{k}}={\frac {n\times (n-1)\times \cdots \times (n-k+1)}{k\times (k-1)\times \cdots \times 1}},}$

which using factorial notation can be compactly expressed as

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

For example, the fourth power of 1 + x is

${\displaystyle {\begin{aligned}(1+x)^{4}&={\tbinom {4}{0}}x^{0}+{\tbinom {4}{1}}x^{1}+{\tbinom {4}{2}}x^{2}+{\tbinom {4}{3}}x^{3}+{\tbinom {4}{4}}x^{4}\\&=1+4x+6x^{2}+4x^{3}+x^{4},\end{aligned}}}$

and the binomial coefficient ${\displaystyle {\tbinom {4}{2}}={\tfrac {4\times 3}{2\times 1}}={\tfrac {4!}{2!2!}}=6}$ is the coefficient of the x² term.

Arranging the numbers ${\displaystyle {\tbinom {n}{0}},{\tbinom {n}{1}},\ldots ,{\tbinom {n}{n}}}$ in successive rows for n = 0, 1, 2, ... gives a triangular array called Pascal's triangle, satisfying the recurrence relation

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

The binomial coefficients occur in many areas of mathematics, and especially in combinatorics. The symbol ${\displaystyle {\tbinom {n}{k}}}$ is usually read as "n choose k" because there are ${\displaystyle {\tbinom {n}{k}}}$ ways to choose an (unordered) subset of k elements from a fixed set of n elements. For example, there are ${\displaystyle {\tbinom {4}{2}}=6}$ 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 ${\displaystyle {\tbinom {z}{k}}}$ for any complex number z and integer k ≥ 0, and many of their properties continue to hold in this more general form.

Andreas von Ettingshausen introduced the notation ${\displaystyle {\tbinom {n}{k}}}$  in 1826, 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.

Alternative notations include C(n, k), _nC_k, ⁿC_k, C^k
_n, Cⁿ
_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.

For natural numbers (taken to include 0) n and k, the binomial coefficient ${\displaystyle {\tbinom {n}{k}}}$  can be defined as the coefficient of the monomial X^k in the expansion of (1 + X)ⁿ. The same coefficient also occurs (if k ≤ n) in the binomial formula

${\displaystyle (x+y)^{n}=\sum _{k=0}^{n}{\binom {n}{k}}x^{k}y^{n-k}}$

()

(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 X^k, and the coefficient of that monomial in the result will be the number of such subsets. This shows in particular that ${\displaystyle {\tbinom {n}{k}}}$  is a natural number for any natural numbers n and k. There are many other combinatorial interpretations of binomial coefficients (counting problems for which the answer is given by a binomial coefficient expression), for instance the number of words formed of n bits (digits 0 or 1) whose sum is k is given by ${\displaystyle {\tbinom {n}{k}}}$ , while the number of ways to write ${\displaystyle k=a_{1}+a_{2}+\cdots +a_{n}}$  where every a_i is a nonnegative integer is given by ${\displaystyle {\tbinom {n+k-1}{n-1}}}$ . Most of these interpretations can be shown to be equivalent to counting k-combinations.

Several methods exist to compute the value of ${\displaystyle {\tbinom {n}{k}}}$  without actually expanding a binomial power or counting k-combinations.

Recursive formula

One method uses the recursive, purely additive formula

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

 

${\displaystyle n,k}$ 

${\displaystyle 1\leq k$ 

$${\displaystyle {\binom {n}{0}}={\binom {n}{n}}=1}$$

 

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 Xⁿ⁺¹ or X⁻¹ in (1 + X)ⁿ, one might extend the definition beyond the above boundaries to include ${\displaystyle {\tbinom {n}{k}}=0}$  when either k > n or k < 0. This recursive formula then allows the construction of Pascal's triangle, surrounded by white spaces where the zeros, or the trivial coefficients, would be.

Multiplicative formula

A more efficient method to compute individual binomial coefficients is given by the formula

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

 

${\displaystyle n^{\underline {k}}}$ 

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 formula

Finally, though computationally unsuitable, there is the compact form, often used in proofs and derivations, which makes repeated use of the familiar factorial function:

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

 

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

()

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

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

 

Generalization and connection to the binomial series

The multiplicative formula allows the definition of binomial coefficients to be extended by replacing n by an arbitrary number α (negative, real, complex) or even an element of any commutative ring in which all positive integers are invertible:

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

 

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

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

()

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

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

 

If α is a nonnegative integer n, then all terms with k > n are zero, and the infinite series becomes a finite sum, thereby recovering the binomial formula. However, for other values of α, including negative integers and rational numbers, the series is really infinite.

 

Pascal's rule is the important recurrence relation

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

()

which can be used to prove by mathematical induction that ${\displaystyle {\tbinom {n}{k}}}$  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 ${\displaystyle {\tbinom {n}{k}}}$  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

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

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

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

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

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

this presents a polynomial in t with rational coefficients.

As such, it can be evaluated at any real or complex number t to define binomial coefficients with such first arguments. These "generalized binomial coefficients" appear in Newton's generalized binomial theorem.

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

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

${\displaystyle {\binom {t}{k}}=\sum _{i=0}^{k}s(k,i){\frac {t^{i}}{k!}}.}$ 

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

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

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

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

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 ${\textstyle \sum _{k=0}^{d}a_{k}{\binom {t}{k}}}$  of binomial coefficients. The coefficient a_k is the kth difference of the sequence p(0), p(1), ..., p(k). Explicitly,

${\displaystyle a_{k}=\sum _{i=0}^{k}(-1)^{k-i}{\binom {k}{i}}p(i).}$

()

Integer-valued polynomials

Each polynomial ${\displaystyle {\tbinom {t}{k}}}$  is integer-valued: it has an integer value at all integer inputs ${\displaystyle t}$ . (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.

Example

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

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

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

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

()

and, with a little more work,

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

We can also get

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

Moreover, the following may be useful:

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

For constant n, we have the following recurrence:

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

To sum up, we have

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

Sums of the binomial coefficients

The formula

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

()

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 ${\displaystyle 2^{n}}$  choices. The left and right sides are two ways to count the same collection of subsets, so they are equal.

The formulas

${\displaystyle \sum _{k=0}^{n}k{\binom {n}{k}}=n2^{n-1}}$

()

and

${\displaystyle \sum _{k=0}^{n}k^{2}{\binom {n}{k}}=(n+n^{2})2^{n-2}}$ 

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

${\displaystyle \sum _{j=0}^{k}{\binom {m}{j}}{\binom {n-m}{k-j}}={\binom {n}{k}}}$

()

and can be found by examination of the coefficient of ${\displaystyle x^{k}}$  in the expansion of (1 + x)^m(1 + x)^n−m = (1 + x)ⁿ using equation (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)

${\displaystyle {\begin{array}{c}1\\1\qquad 1\\1\qquad 2\qquad 1\\{\color {blue}1\qquad 3\qquad 3\qquad 1}\\1\qquad 4\qquad 6\qquad 4\qquad 1\\1\qquad 5\qquad 10\qquad 10\qquad 5\qquad 1\\1\qquad 6\qquad 15\qquad {\color {red}20}\qquad 15\qquad 6\qquad 1\\1\qquad 7\qquad 21\qquad 35\qquad 35\qquad 21\qquad 7\qquad 1\end{array}}}$ 

Pascal's triangle, rows 0 through 7. Equation 8 for m = 3 is illustrated in rows 3 and 6 as ${\displaystyle 1^{2}+3^{2}+3^{2}+1^{2}=20.}$ 

${\displaystyle \sum _{j=0}^{m}{\binom {m}{j}}^{2}={\binom {2m}{m}},}$

()

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 0 ≤ j ≤ k ≤ n, is

${\displaystyle \sum _{m=0}^{n}{\binom {m}{j}}{\binom {n-m}{k-j}}={\binom {n+1}{k+1}}.}$

()

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

${\displaystyle \sum _{m=k}^{n}{\binom {m}{k}}={\binom {n+1}{k+1}}}$ 

and its relative

${\displaystyle \sum _{r=0}^{m}{\binom {n+r}{r}}={\binom {n+m+1}{m}}.}$ 

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

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

This can be proved by induction using (3) or by Zeckendorf's representation. A combinatorial proof is given below.

Multisections of sums

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

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

For small s, these series have particularly nice forms; for example,

${\displaystyle {\binom {n}{0}}+{\binom {n}{3}}+{\binom {n}{6}}+\cdots ={\frac {1}{3}}\left(2^{n}+2\cos {\frac {n\pi }{3}}\right)}$ 
${\displaystyle {\binom {n}{1}}+{\binom {n}{4}}+{\binom {n}{7}}+\cdots ={\frac {1}{3}}\left(2^{n}+2\cos {\frac {(n-2)\pi }{3}}\right)}$ 
${\displaystyle {\binom {n}{2}}+{\binom {n}{5}}+{\binom {n}{8}}+\cdots ={\frac {1}{3}}\left(2^{n}+2\cos {\frac {(n-4)\pi }{3}}\right)}$ 
${\displaystyle {\binom {n}{0}}+{\binom {n}{4}}+{\binom {n}{8}}+\cdots ={\frac {1}{2}}\left(2^{n-1}+2^{\frac {n}{2}}\cos {\frac {n\pi }{4}}\right)}$ 
${\displaystyle {\binom {n}{1}}+{\binom {n}{5}}+{\binom {n}{9}}+\cdots ={\frac {1}{2}}\left(2^{n-1}+2^{\frac {n}{2}}\sin {\frac {n\pi }{4}}\right)}$ 
${\displaystyle {\binom {n}{2}}+{\binom {n}{6}}+{\binom {n}{10}}+\cdots ={\frac {1}{2}}\left(2^{n-1}-2^{\frac {n}{2}}\cos {\frac {n\pi }{4}}\right)}$ 
${\displaystyle {\binom {n}{3}}+{\binom {n}{7}}+{\binom {n}{11}}+\cdots ={\frac {1}{2}}\left(2^{n-1}-2^{\frac {n}{2}}\sin {\frac {n\pi }{4}}\right)}$ 

Partial sums

Although there is no closed formula for partial sums

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

of binomial coefficients, one can again use (3) and induction to show that for k = 0, …, n − 1,

${\displaystyle \sum _{j=0}^{k}(-1)^{j}{\binom {n}{j}}=(-1)^{k}{\binom {n-1}{k}},}$ 

with special case

${\displaystyle \sum _{j=0}^{n}(-1)^{j}{\binom {n}{j}}=0}$ 

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,

${\displaystyle \sum _{j=0}^{n}(-1)^{j}{\binom {n}{j}}P(j)=0.}$ 

Differentiating (2) k times and setting x = −1 yields this for ${\displaystyle P(x)=x(x-1)\cdots (x-k+1)}$ , when 0 ≤ k < n, and the general case follows by taking linear combinations of these.

When P(x) is of degree less than or equal to n,

${\displaystyle \sum _{j=0}^{n}(-1)^{j}{\binom {n}{j}}P(n-j)=n!a_{n}}$

()

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

More generally for (10),

${\displaystyle \sum _{j=0}^{n}(-1)^{j}{\binom {n}{j}}P(m+(n-j)d)=d^{n}n!a_{n}}$ 

where m and d are complex numbers. This follows immediately applying (10) to the polynomial ${\displaystyle Q(x):=P(m+dx)}$  instead of ${\displaystyle P(x)}$ , and observing that ${\displaystyle Q(x)}$  still has degree less than or equal to n, and that its coefficient of degree n is dⁿa_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 proofs

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

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

(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 ${\displaystyle {\tbinom {n}{q}}}$  ways of choosing a set of q elements to mark, and ${\displaystyle 2^{n-q}}$  to choose which of the remaining elements of [n] also belong to the subset.

In Pascal's identity

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

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

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

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

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

Now apply (1) to get the result.

If one denotes by F(i) the sequence of Fibonacci numbers, indexed so that F(0) = F(1) = 1, then the identity

$${\displaystyle \sum _{k=0}^{\left\lfloor {\frac {n}{2}}\right\rfloor }{\binom {n-k}{k}}=F(n)}$$

 

${\displaystyle {\tbinom {n-k}{k}}}$ 

Sum of coefficients row

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 ${\displaystyle 2^{n}-1}$ , where each digit position is an item from the set of n.

Dixon's identity

Dixon's identity is

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

or, more generally,

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

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

Continuous identities

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

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

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

$${\displaystyle {\binom {n-1}{k}}\equiv (-1)^{k}\mod n}$$

 

${\displaystyle 0\leq k\leq n-1.}$ 

Indeed, we have

${\displaystyle {\binom {n-1}{k}}={(n-1)(n-2)\cdots (n-k) \over 1\cdot 2\cdots k}=\prod _{i=1}^{k}{n-i \over i}\equiv \prod _{i=1}^{k}{-i \over i}=(-1)^{k}\mod n.}$ 

Ordinary generating functions

For a fixed n, the ordinary generating function of the sequence ${\displaystyle {\tbinom {n}{0}},{\tbinom {n}{1}},{\tbinom {n}{2}},\ldots }$  is

${\displaystyle \sum _{k=0}^{\infty }{n \choose k}x^{k}=(1+x)^{n}.}$ 

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

${\displaystyle \sum _{n=0}^{\infty }{n \choose k}y^{n}={\frac {y^{k}}{(1-y)^{k+1}}}.}$ 

The bivariate generating function of the binomial coefficients is

${\displaystyle \sum _{n=0}^{\infty }\sum _{k=0}^{n}{n \choose k}x^{k}y^{n}={\frac {1}{1-y-xy}}.}$ 

A symmetric bivariate generating function of the binomial coefficients is

${\displaystyle \sum _{n=0}^{\infty }\sum _{k=0}^{\infty }{n+k \choose k}x^{k}y^{n}={\frac {1}{1-x-y}}.}$ 

which is the same as the previous generating function after the substitution ${\displaystyle x\to xy}$ .

Exponential generating function

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

${\displaystyle \sum _{n=0}^{\infty }\sum _{k=0}^{\infty }{n+k \choose k}{\frac {x^{k}y^{n}}{(n+k)!}}=e^{x+y}.}$ 

In 1852, Kummer proved that if m and n are nonnegative integers and p is a prime number, then the largest power of p dividing ${\displaystyle {\tbinom {m+n}{m}}}$  equals 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 ${\displaystyle {\tbinom {n}{k}}}$  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 ${\displaystyle {\tbinom {n}{k}}}$  is divisible by n/gcd(n,k). In particular therefore it follows that p divides ${\displaystyle {\tbinom {p^{r}}{s}}}$  for all positive integers r and s such that s < p^r. However this is not true of higher powers of p: for example 9 does not divide ${\displaystyle {\tbinom {9}{6}}}$ .

A somewhat surprising result by David Singmaster (1974) is that any integer divides almost all binomial coefficients. More precisely, fix an integer d and let f(N) denote the number of binomial coefficients ${\displaystyle {\tbinom {n}{k}}}$  with n < N such that d divides ${\displaystyle {\tbinom {n}{k}}}$ . Then

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

Since the number of binomial coefficients ${\displaystyle {\tbinom {n}{k}}}$  with n < N is N(N + 1) / 2, this implies that the density of binomial coefficients divisible by d goes to 1.

Binomial coefficients have divisibility properties related to least common multiples of consecutive integers. For example:

${\displaystyle {\binom {n+k}{k}}}$  divides ${\displaystyle {\frac {\operatorname {lcm} (n,n+1,\ldots ,n+k)}{n}}}$ .
${\displaystyle {\binom {n+k}{k}}}$  is a multiple of ${\displaystyle {\frac {\operatorname {lcm} (n,n+1,\ldots ,n+k)}{n\cdot \operatorname {lcm} ({\binom {k}{0}},{\binom {k}{1}},\ldots ,{\binom {k}{k}})}}}$ .

Another fact: An integer n ≥ 2 is prime if and only if all the intermediate binomial coefficients

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

are divisible by n.

Proof: When p is prime, p divides

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

because ${\displaystyle {\tbinom {p}{k}}}$  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

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

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

The following bounds for ${\displaystyle {\tbinom {n}{k}}}$  hold for all values of n and k such that 1 ≤ k ≤ n:

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

 

$${\displaystyle {n \choose k}={\frac {n}{k}}\cdot {\frac {n-1}{k-1}}\cdots {\frac {n-(k-1)}{1}}}$$

 

${\displaystyle k}$ 

${\textstyle \geq {\frac {n}{k}}}$ 

${\textstyle e^{k}>k^{k}/k!}$ 

${\textstyle e^{k}=\sum _{j=0}^{\infty }k^{j}/j!}$ 

From the divisibility properties we can infer that

$${\displaystyle {\frac {\operatorname {lcm} (n-k,\ldots ,n)}{(n-k)\cdot \operatorname {lcm} \left({\binom {k}{0}},\ldots ,{\binom {k}{k}}\right)}}\leq {\binom {n}{k}}\leq {\frac {\operatorname {lcm} (n-k,\ldots ,n)}{n-k}},}$$

 

The following bounds are useful in information theory:^: 353

$${\displaystyle {\frac {1}{n+1}}2^{nH(k/n)}\leq {n \choose k}\leq 2^{nH(k/n)}}$$

 

${\displaystyle H(p)=-p\log _{2}(p)-(1-p)\log _{2}(1-p)}$ 

$${\displaystyle {\sqrt {\frac {n}{8k(n-k)}}}2^{nH(k/n)}\leq {n \choose k}\leq {\sqrt {\frac {n}{2\pi k(n-k)}}}2^{nH(k/n)}}$$

 

${\displaystyle 1\leq k\leq n-1}$ 

Both n and k large

Stirling's approximation yields the following approximation, valid when ${\displaystyle n-k,k}$  both tend to infinity:

$${\displaystyle {n \choose k}\sim {\sqrt {n \over 2\pi k(n-k)}}\cdot {n^{n} \over k^{k}(n-k)^{n-k}}}$$

 

${\displaystyle n}$ 

$${\displaystyle {2n \choose n}\sim {\frac {2^{2n}}{\sqrt {n\pi }}}}$$

 

${\displaystyle {\sqrt {n}}{2n \choose n}\geq 2^{2n-1}}$ 

$${\displaystyle {\sqrt {n}}{mn \choose n}\geq {\frac {m^{m(n-1)+1}}{(m-1)^{(m-1)(n-1)}}}.}$$

 

If n is large and k is linear in n, various precise asymptotic estimates exist for the binomial coefficient ${\textstyle {\binom {n}{k}}}$ . For example, if ${\displaystyle |n/2-k|=o(n^{2/3})}$  then

$${\displaystyle {\binom {n}{k}}\sim {\binom {n}{\frac {n}{2}}}e^{-d^{2}/(2n)}\sim {\frac {2^{n}}{\sqrt {{\frac {1}{2}}n\pi }}}e^{-d^{2}/(2n)}}$$

 

n much larger than k

If n is large and k is o(n) (that is, if k/n → 0), then

$${\displaystyle {\binom {n}{k}}\sim \left({\frac {ne}{k}}\right)^{k}\cdot (2\pi k)^{-1/2}\cdot \exp \left(-{\frac {k^{2}}{2n}}(1+o(1))\right)}$$

 

Sums of binomial coefficients

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

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