Binomial coefficient#Generalization to multinomials

Andreas von Ettingshausen introduced the notation $\backslash tbinom\; nk$ in 1826,{{harvtxt|Higham|1998}} 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.Lilavati Section 6, Chapter 4 (see {{harvtxt|Knuth|1997}}).

Alternative notations include {{math|C(n, k)}}, {{math|{{sub|n}}C{{sub|k}}}}, {{math|{{sup|n}}C_k}}, {{math|C{{su|p=k|b=n|lh=0.8}}}},{{harvnb|Uspensky|1937|p=18}} {{math|C{{su|p=n|b=k|lh=0.8}}}}, and {{math|C{{sub|n,k}}}}, in all of which the {{mvar|C}} stands for combinations or choices; the {{mvar|C}} notation means the number of ways to choose k out of n objects. Many calculators use variants of the {{nowrap|{{mvar|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 permutation#k-permutations of n, written as {{math|P(n, k)}}, etc.

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

{{NumBlk2|:|$(x+y)^n=\backslash sum\_\{k=0\}^n\backslash binom\; nk\; x^ky^\{n-k\}$|∗}}

(valid for any elements {{mvar|x}}, {{mvar|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 {{mvar|k}} objects can be chosen from among {{mvar|n}} objects; more formally, the number of {{mvar|k}}-element subsets (or {{mvar|k}}-combinations) of an {{mvar|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 {{mvar|n}} factors of the power {{math|(1 + X)ⁿ}} one temporarily labels the term {{mvar|X}} with an index {{mvar|i}} (running from {{math|1}} to {{mvar|n}}), then each subset of {{mvar|k}} indices gives after expansion a contribution {{math|X^k}}, and the coefficient of that monomial in the result will be the number of such subsets. This shows in particular that $\backslash tbinom\; nk$ is a natural number for any natural numbers {{mvar|n}} and {{mvar|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 {{mvar|n}} bits (digits 0 or 1) whose sum is {{mvar|k}} is given by $\backslash tbinom\; nk$, while the number of ways to write $k\; =\; a\_1\; +\; a\_2\; +\; \backslash cdots\; +\; a\_n$ where every {{math|a_i}} is a nonnegative integer is given by {{tmath|1= \tbinom{n+k-1}{n-1} }}. Most of these interpretations can be shown to be equivalent to counting {{mvar|k}}-combinations.

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

One method uses the recursive, purely additive formula

$$\backslash binom\; nk\; =\; \backslash binom\{n-1\}\{k-1\}\; +\; \backslash binom\{n-1\}k$$ for all integers $n,k$ such that

with boundary values

$$\backslash binom\; n0\; =\; \backslash binom\; nn\; =\; 1$$

for all integers {{math|1=n ≥ 0}}.

The formula follows from considering the set {{math|{{mset|1, 2, 3, ..., n}}}} and counting separately (a) the {{mvar|k}}-element groupings that include a particular set element, say "{{mvar|i}}", in every group (since "{{mvar|i}}" is already chosen to fill one spot in every group, we need only choose {{math|k − 1}} from the remaining {{math|n − 1}}) and (b) all the k-groupings that don't include "{{mvar|i}}"; this enumerates all the possible {{mvar|k}}-combinations of {{mvar|n}} elements. It also follows from tracing the contributions to X^k in {{math|(1 + X)ⁿ⁻¹(1 + X)}}. As there is zero {{math|Xⁿ⁺¹}} or {{math|X⁻¹}} in {{math|(1 + X)ⁿ}}, one might extend the definition beyond the above boundaries to include $\backslash tbinom\; nk\; =\; 0$ when either {{math|k > n}} or {{math|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

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

where the numerator of the first fraction, $n^\{\backslash underline\{k\}\}$, 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 {{mvar|k}} distinct objects, retaining the order of selection, from a set of {{mvar|n}} objects. The denominator counts the number of distinct sequences that define the same {{mvar|k}}-combination when order is disregarded. This formula can also be stated in a recursive form. Using the "C" notation from above, $C\_\{n,k\}\; =\; C\_\{n,\; k-1\}\; \backslash cdot\; (n-k+1)\; /\; k$, 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 {{mvar|k}} and {{math|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 {{mvar|k}} and {{math|n − k}}.

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

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

where {{math|n!}} denotes the factorial of {{mvar|n}}. This formula follows from the multiplicative formula above by multiplying numerator and denominator by {{math|(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 {{mvar|k}} is small and {{mvar|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)

{{NumBlk2|:|$\backslash binom\; nk\; =\; \backslash binom\; n\{n-k\}\; \backslash quad\; \backslash text\{for\; \}\backslash \; 0\backslash leq\; k\backslash leq\; n,$|1}}

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

$$\backslash binom\; nk\; =$$

\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}.

{{Main|Binomial series}}

The multiplicative formula allows the definition of binomial coefficients to be extendedSee {{Harv|Graham|Knuth|Patashnik|1994}}, which also defines $\backslash tbinom\; n\; k\; =\; 0$ for . Alternative generalizations, such as to two real or complex valued arguments using the Gamma function assign nonzero values to $\backslash tbinom\; n\; k$ for , 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). 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:

$$\backslash binom\; \backslash alpha\; k\; =\; \backslash frac\{\backslash alpha^\{\backslash underline\; k\}\}\{k!\}\; =\; \backslash frac\{\backslash alpha(\backslash alpha-1)(\backslash alpha-2)\backslash cdots(\backslash alpha-k+1)\}\{k(k-1)(k-2)\backslash cdots\; 1\}$$

\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 $\backslash tbinom\backslash alpha\; k$ binomial coefficients:

{{NumBlk2|:|$(1+X)^\backslash alpha\; =\; \backslash sum\_\{k=0\}^\backslash infty\; \{\backslash alpha\; \backslash choose\; k\}\; X^k.$|2}}

This formula is valid for all complex numbers α and X with |X| < 1. It can also be interpreted as an identity of formal power series in X, where it actually can serve as definition of arbitrary powers of power series with constant coefficient equal to 1; the point is that with this definition all identities hold that one expects for exponentiation, notably

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

If α is a nonnegative integer n, then all terms with {{math|k > n}} are zero,When $\backslash alpha\; =\; n$ is a nonnegative integer, $\backslash textstyle\; \backslash binom\{n\}\{k\}\; =\; 0$ 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 $k\; \backslash geq\; n\; +\; 1$. 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.

Image:Pascal's triangle - 1000th row.png.]]

{{Main|Pascal's triangle|Pascal's rule}}

Pascal's rule is the important recurrence relation

{{NumBlk2|:|$\{n\; \backslash choose\; k\}\; +\; \{n\; \backslash choose\; k+1\}\; =\; \{n+1\; \backslash choose\; k+1\},$|3}}

which can be used to prove by mathematical induction that $\backslash tbinom\; n\; k$ 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 {{mvar|n}} contains the numbers $\backslash tbinom\{n\}\{k\}$ for {{math|1=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

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

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

• There are $\backslash tbinom\; n\; k$ ways to choose k elements from a set of n elements. See Combination.
• There are $\backslash tbinom\; \{n+k-1\}k$ ways to choose k elements from a set of n elements if repetitions are allowed. See Multiset.
• There are $\backslash tbinom\; \{n+k\}\; k$ strings containing k ones and n zeros.
• There are $\backslash tbinom\; \{n+1\}\; k$ strings consisting of k ones and n zeros such that no two ones are adjacent.{{cite journal|last=Muir|first=Thomas|title=Note on Selected Combinations|journal=Proceedings of the Royal Society of Edinburgh|year=1902|url=https://books.google.com/books?id=EN8vAAAAIAAJ&pg=GBS.PA102}}
• The Catalan numbers are $\backslash tfrac\{1\}\{n+1\}\backslash tbinom\{2n\}\{n\}.$
• The binomial distribution in statistics is $\backslash tbinom\; n\; k\; p^k\; (1-p)^\{n-k\}\; .$

For any nonnegative integer k, the expression $\backslash binom\{t\}\{k\}$ can be written as a polynomial with denominator {{math|k!}}:

: $\backslash binom\{t\}\{k\}\; =\; \backslash frac\{t^\backslash underline\{k\}\}\{k!\}\; =\; \backslash frac\{t(t-1)(t-2)\backslash cdots(t-k+1)\}\{k(k-1)(k-2)\backslash cdots2\; \backslash cdot\; 1\};$

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 $\backslash tbinom\{t\}\{k\}$ can be characterized as the unique degree k polynomial {{math|p(t)}} satisfying {{math|1=p(0) = p(1) = ⋯ = p(k − 1) = 0}} and {{math|1=p(k) = 1}}.

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

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

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

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

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

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

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 $\backslash sum\_\{k=0\}^d\; a\_k\; \backslash binom\{t\}\{k\}$ 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,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.

{{NumBlk2|:|$a\_k\; =\; \backslash sum\_\{i=0\}^k\; (-1)^\{k-i\}\; \backslash binom\{k\}\{i\}\; p(i).$|4}}

{{main|Integer-valued polynomial}}

Each polynomial $\backslash tbinom\{t\}\{k\}$ 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, ({{EquationNote|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 integer-valued polynomial {{math|3t(3t + 1) / 2}} can be rewritten as

:$9\backslash binom\{t\}\{2\}\; +\; 6\; \backslash binom\{t\}\{1\}\; +\; 0\backslash binom\{t\}\{0\}.$

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

{{NumBlk2|:|$\backslash binom\{n\}\{k\}\; =\; \backslash frac\{n\}\{k\}\; \backslash binom\{n-1\}\{k-1\}$|5}}

and, with a little more work,

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

We can also get

: $\backslash binom\; \{n-1\}\{k\}\; =\; \backslash frac\{n-k\}\{n\}\; \backslash binom\; \{n\}\{k\}.$

Moreover, the following may be useful:

:$\backslash binom\{n\}\{k\}\backslash binom\{k\}\{j\}\; =\; \backslash binom\{n\}\{j\}\backslash binom\{n-j\}\{k-j\}=\backslash binom\{n\}\{k-j\}\backslash binom\{n-k+j\}\{j\}.$

For constant n, we have the following recurrence:

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

To sum up, we have

: $\backslash binom\; \{n\}\{k\}\; =\; \backslash binom\; n\{n-k\}\; =\; \backslash frac\{n-k+1\}\{k\}\; \backslash binom\; \{n\}\{k-1\}\; =\; \backslash frac\{n\}\{n-k\}\; \backslash binom\; \{n-1\}\{k\}$

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

The formula

{{NumBlk2|:|$\backslash sum\_\{k=0\}^n\; \backslash binom\; n\; k\; =\; 2^n$|∗∗}}

says that the elements in the {{mvar|n}}th row of Pascal's triangle always add up to 2 raised to the {{mvar|n}}th power. This is obtained from the binomial theorem ({{EquationNote|∗}}) by setting {{math|1=x = 1}} and {{math|1=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

{{NumBlk2|:|$\backslash sum\_\{k=0\}^n\; k\; \backslash binom\{n\}\{k\}\; =\; n\; 2^\{n-1\}$|6}}

and

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

follow from the binomial theorem after differentiating with respect to {{mvar|x}} (twice for the latter) and then substituting {{math|1=x = y = 1}}.

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

{{NumBlk2|:|$\backslash sum\_\{j=0\}^k\; \backslash binom\; m\; j\; \backslash binom\{n-m\}\{k-j\}\; =\; \backslash binom\; n\; k$|7}}

and can be found by examination of the coefficient of $x^k$ in the expansion of {{math|1=(1 + x)^m(1 + x)^n−m = (1 + x)ⁿ}} using equation ({{EquationNote|2}}). When {{math|1=m = 1}}, equation ({{EquationNote|7}}) reduces to equation ({{EquationNote|3}}). In the special case {{math|1=n = 2m, k = m}}, using ({{EquationNote|1}}), the expansion ({{EquationNote|7}}) becomes (as seen in Pascal's triangle at right)

{{Image frame|width=395

|content=

\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}

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

}}

{{NumBlk2|:|$\backslash sum\_\{j=0\}^m\; \backslash binom\{m\}\{j\}\; ^2\; =\; \backslash binom\; \{2m\}\; m,$|8}}

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 {{math|1=0 ≤ j ≤ k ≤ n}}, is

{{NumBlk2|:|$\backslash sum\_\{m=0\}^n\; \backslash binom\; m\; j\; \backslash binom\; \{n-m\}\{k-j\}=\; \backslash binom\; \{n+1\}\{k+1\}.$|9}}

The proof is similar, but uses the binomial series expansion ({{EquationNote|2}}) with negative integer exponents.

When {{math|1=j = k}}, equation ({{EquationNote|9}}) gives the hockey-stick identity

: $\backslash sum\_\{m=k\}^n\; \backslash binom\; m\; k\; =\; \backslash binom\; \{n+1\}\{k+1\}$

and its relative

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

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

Then

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

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

For integers s and t such that series multisection gives the following identity for the sum of binomial coefficients:

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

For small {{mvar|s}}, these series have particularly nice forms; for example,{{harvtxt|Gradshteyn|Ryzhik|2014|pp=3–4}}.

: $\backslash binom\{n\}\{0\}\; +\; \backslash binom\{n\}\{3\}+\backslash binom\{n\}\{6\}+\backslash cdots\; =\; \backslash frac\{1\}\{3\}\backslash left(2^n\; +2\; \backslash cos\backslash frac\{n\backslash pi\}\{3\}\backslash right)$

: $\backslash binom\{n\}\{1\}\; +\; \backslash binom\{n\}\{4\}+\backslash binom\{n\}\{7\}+\backslash cdots\; =\; \backslash frac\{1\}\{3\}\backslash left(2^n\; +2\; \backslash cos\backslash frac\{(n-2)\backslash pi\}\{3\}\backslash right)$

: $\backslash binom\{n\}\{2\}\; +\; \backslash binom\{n\}\{5\}+\backslash binom\{n\}\{8\}+\backslash cdots\; =\; \backslash frac\{1\}\{3\}\backslash left(2^n\; +2\; \backslash cos\backslash frac\{(n-4)\backslash pi\}\{3\}\backslash right)$

: $\backslash binom\{n\}\{0\}\; +\; \backslash binom\{n\}\{4\}+\backslash binom\{n\}\{8\}+\backslash cdots\; =\; \backslash frac\{1\}\{2\}\backslash left(2^\{n-1\}\; +2^\{\backslash frac\{n\}\{2\}\}\; \backslash cos\backslash frac\{n\backslash pi\}\{4\}\backslash right)$

: $\backslash binom\{n\}\{1\}\; +\; \backslash binom\{n\}\{5\}+\backslash binom\{n\}\{9\}+\backslash cdots\; =\; \backslash frac\{1\}\{2\}\backslash left(2^\{n-1\}\; +2^\{\backslash frac\{n\}\{2\}\}\; \backslash sin\backslash frac\{n\backslash pi\}\{4\}\backslash right)$

: $\backslash binom\{n\}\{2\}\; +\; \backslash binom\{n\}\{6\}+\backslash binom\{n\}\{10\}+\backslash cdots\; =\; \backslash frac\{1\}\{2\}\backslash left(2^\{n-1\}\; -2^\{\backslash frac\{n\}\{2\}\}\; \backslash cos\backslash frac\{n\backslash pi\}\{4\}\backslash right)$

: $\backslash binom\{n\}\{3\}\; +\; \backslash binom\{n\}\{7\}+\backslash binom\{n\}\{11\}+\backslash cdots\; =\; \backslash frac\{1\}\{2\}\backslash left(2^\{n-1\}\; -2^\{\backslash frac\{n\}\{2\}\}\; \backslash sin\backslash frac\{n\backslash pi\}\{4\}\backslash right)$

Although there is no closed formula for partial sums

: $\backslash sum\_\{j=0\}^k\; \backslash binom\; n\; j$

of binomial coefficients,{{citation | last = Boardman | first = Michael | issue = 5 | journal = Mathematics Magazine | jstor = 3219201 | doi = 10.2307/3219201 | mr = 1573776 | quote = it is well known that there is no closed form (that is, direct formula) for the partial sum of binomial coefficients | pages = 368–372 | title = The Egg-Drop Numbers | volume = 77 | year = 2004}}. one can again use ({{EquationNote|3}}) and induction to show that for {{math|1=k = 0, …, n − 1}},

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

with special casesee induction developed in eq (7) p. 1389 in {{citation

| last = Aupetit | first = Michael

| journal = Neurocomputing

| pages = 1379–1389

| title = Nearly homogeneous multi-partitioning with a deterministic generator

| volume = 72

| number = 7–9

| year = 2009

| issn = 0925-2312

| doi = 10.1016/j.neucom.2008.12.024

}}.

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

for {{math|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,{{cite journal|last=Ruiz|first=Sebastian|title=An Algebraic Identity Leading to Wilson's Theorem|journal=The Mathematical Gazette|year=1996|volume=80|issue=489|pages=579–582|doi=10.2307/3618534| jstor=3618534|arxiv=math/0406086|s2cid=125556648 }}

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

Differentiating ({{EquationNote|2}}) k times and setting x = −1 yields this for

$P(x)=x(x-1)\backslash cdots(x-k+1)$,

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,

{{NumBlk2|:|$\backslash sum\_\{j=0\}^n\; (-1)^j\backslash binom\; n\; j\; P(n-j)\; =\; n!a\_n$|10}}

where $a\_n$ is the coefficient of degree n in P(x).

More generally for ({{EquationNote|10}}),

: $\backslash sum\_\{j=0\}^n\; (-1)^j\backslash binom\; n\; j\; P(m+(n-j)d)\; =\; d^n\; n!\; a\_n$

where m and d are complex numbers. This follows immediately applying ({{EquationNote|10}}) to the polynomial {{tmath|1=Q(x):=P(m + dx)}} instead of {{tmath|P(x)}}, and observing that {{tmath|Q(x)}} still has degree less than or equal to n, and that its coefficient of degree n is dⁿa_n.

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

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

: $\backslash sum\_\{k=q\}^n\; \backslash binom\{n\}\{k\}\; \backslash binom\{k\}\{q\}\; =\; 2^\{n-q\}\backslash binom\{n\}\{q\}$

(which reduces to ({{EquationNote|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 $\backslash tbinom\; n\; q$ 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.

In Pascal's identity

: $\{n\; \backslash choose\; k\}\; =\; \{n-1\; \backslash choose\; k-1\}\; +\; \{n-1\; \backslash choose\; k\},$

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 ({{EquationNote|8}}) also has a combinatorial proof. The identity reads

: $\backslash sum\_\{k=0\}^n\; \backslash binom\{n\}\{k\}^2\; =\; \backslash binom\{2n\}\{n\}.$

Suppose you have $2n$ empty squares arranged in a row and you want to mark (select) n of them. There are $\backslash tbinom\; \{2n\}n$ 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

:$\backslash sum\_\{k=0\}^n\backslash binom\; n\; k\backslash binom\; n\{n-k\}\; =\; \backslash binom\; \{2n\}\; n.$

Now apply ({{EquationNote|1}}) to get the result.

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

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

has the following combinatorial proof.{{harvnb|Benjamin|Quinn|2003|loc=pp. 4−5}} One may show by induction that {{math|F(n)}} counts the number of ways that a {{math|n × 1}} strip of squares may be covered by {{math|2 × 1}} and {{math|1 × 1}} tiles. On the other hand, if such a tiling uses exactly {{mvar|k}} of the {{math|2 × 1}} tiles, then it uses {{math|n − 2k}} of the {{math|1 × 1}} tiles, and so uses {{math|n − k}} tiles total. There are $\backslash tbinom\{n-k\}\{k\}$ ways to order these tiles, and so summing this coefficient over all possible values of {{mvar|k}} gives the identity.

{{See also|Combination#Number of k-combinations for all k}}

The number of k-combinations for all k, $\backslash sum\_\{0\backslash leq\{k\}\backslash leq\{n\}\}\backslash binom\; nk\; =\; 2^n$, is the sum of the nth row (counting from 0) of the binomial coefficients. These combinations are enumerated by the 1 digits of the set of base 2 numbers counting from 0 to $2^n\; -\; 1$, where each digit position is an item from the set of n.

Dixon's identity is

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

or, more generally,

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

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

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

: $\backslash int\_\{-\backslash pi\}^\{\backslash pi\}\; \backslash cos((2m-n)x)\backslash cos^n(x)\backslash \; dx\; =\; \backslash frac\{\backslash pi\}\{2^\{n-1\}\}\; \backslash binom\{n\}\{m\}$

: $$

\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}

: $$

\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.

If n is prime, then $$\backslash binom\; \{n-1\}k\; \backslash equiv\; (-1)^k\; \backslash mod\; n$$ for every k with $0\backslash leq\; k\; \backslash leq\; n-1.$

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 = {(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.

For a fixed {{mvar|n}}, the ordinary generating function of the sequence $\backslash tbinom\; n0,\backslash tbinom\; n1,\backslash tbinom\; n2,\backslash ldots$ is

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

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

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

The bivariate generating function of the binomial coefficients is

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

A symmetric bivariate generating function of the binomial coefficients is

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

which is the same as the previous generating function after the substitution $x\backslash to\; xy$.

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

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

{{main|Kummer's theorem|Lucas' theorem}}

In 1852, Kummer proved that if m and n are nonnegative integers and p is a prime number, then the largest power of p dividing $\backslash tbinom\{m+n\}\{m\}$ 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 $\backslash tbinom\; n\; k$

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 $\backslash tbinom\; n\; k$ is divisible by n/gcd(n,k). In particular therefore it follows that p divides $\backslash tbinom\{p^r\}\{s\}$ for all positive integers r and s such that {{math|s < p^r}}. However this is not true of higher powers of p: for example 9 does not divide $\backslash tbinom\{9\}\{6\}$.

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

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

Since the number of binomial coefficients $\backslash tbinom\; n\; k$ 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:{{cite journal |last=Farhi |first=Bakir |title=Nontrivial lower bounds for the least common multiple of some finite sequence of integers |journal=Journal of Number Theory |volume=125 |issue=2 |year=2007 |pages=393–411 |doi=10.1016/j.jnt.2006.10.017 |arxiv=0803.0290 |s2cid=115167580 }}

: $\backslash binom\{n+k\}k$ divides $\backslash frac\{\backslash operatorname\{lcm\}(n,n+1,\backslash ldots,n+k)\}n$.

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

Another fact:

An integer {{math|n ≥ 2}} is prime if and only if

all the intermediate binomial coefficients

: $\backslash binom\; n\; 1,\; \backslash binom\; n\; 2,\; \backslash ldots,\; \backslash binom\; n\{n-1\}$

are divisible by n.

Proof:

When p is prime, p divides

: $\backslash binom\; p\; k\; =\; \backslash frac\{p\; \backslash cdot\; (p-1)\; \backslash cdots\; (p-k+1)\}\{k\; \backslash cdot\; (k-1)\; \backslash cdots\; 1\}$ for all {{math|1=0 < k < p}}

because $\backslash tbinom\; p\; k$ 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 {{math|1=k = n/p}}. Then {{math|1=0 < p < n}} and

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

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

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

The first inequality follows from the fact that

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

and each of these $k$ terms in this product is $\backslash geq\; \backslash frac\{n\}\{k\}$. 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 $e^k\; =\; \backslash sum\_\{j=0\}^\backslash infty\; k^j/j!$.

From the divisibility properties we can infer that

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

where both equalities can be achieved.

The following bounds are useful in information theory:{{cite book |author1=Thomas M. Cover |author2=Joy A. Thomas |title=Elements of Information Theory |date=18 July 2006 |publisher=Wiley |location=Hoboken, New Jersey |isbn=0-471-24195-4}}{{rp|353}}

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

where $H(p)\; =\; -p\backslash log\_2(p)\; -(1-p)\backslash log\_2(1-p)$ is the binary entropy function. It can be further tightened to

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

for all $1\; \backslash leq\; k\; \backslash leq\; n-1$.{{cite book |author1=F. J. MacWilliams |author2=N. J. A. Sloane |title=The Theory of Error-Correcting Codes |volume=16 |edition=3rd |year=1981 |publisher=North-Holland |isbn=0-444-85009-0}}{{rp|309}}

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

$$\{n\; \backslash choose\; k\}\; \backslash sim$$

\sqrt{n\over 2\pi k (n-k)} \cdot

{n^n \over 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

$$\{2n\; \backslash choose\; n\}\; \backslash sim\; \backslash frac\{2^\{2n\}\}\{\backslash sqrt\{n\backslash pi\; \}\}$$ and $\backslash sqrt\{n\}\{2n\; \backslash choose\; n\}\; \backslash ge\; 2^\{2n-1\}$. More generally, for {{math|m ≥ 2}} and {{math|n ≥ 1}} (again, by applying Stirling's formula to the factorials in the binomial coefficient),

$$\backslash sqrt\{n\}\{mn\; \backslash choose\; n\}\; \backslash ge\; \backslash frac\{m^\{m(n-1)+1\}\}\{(m-1)^\{(m-1)(n-1)\}\}.$$

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

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

where d = n − 2k.{{cite book|title=Asymptopia|last1=Spencer|first1=Joel|last2=Florescu|first2=Laura|date=2014| publisher=AMS|isbn=978-1-4704-0904-3|series=Student mathematical library|volume=71|page=66| oclc=865574788|author1-link=Joel Spencer}}

If {{mvar|n}} is large and {{mvar|k}} is {{math|o(n)}} (that is, if {{math| k/n → 0}}), then

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

where again {{mvar|o}} is the little o notation.{{cite book|title=Asymptopia|last1=Spencer|first1=Joel| last2=Florescu|first2=Laura|date=2014|publisher=AMS|isbn=978-1-4704-0904-3|series=Student mathematical library|volume=71|page=59|oclc=865574788|author1-link=Joel Spencer}}

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

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

More precise bounds are given by

$$\backslash frac\{1\}\{\backslash sqrt\{8n\backslash varepsilon(1-\backslash varepsilon)\}\}\; \backslash cdot\; 2^\{H(\backslash varepsilon)\; \backslash cdot\; n\}\; \backslash leq\; \backslash sum\_\{i=0\}^\{k\}\; \backslash binom\{n\}\{i\}\; \backslash leq\; 2^\{H(\backslash varepsilon)\; \backslash cdot\; n\},$$

valid for all integers $n\; >\; k\; \backslash geq\; 1$ with $\backslash varepsilon\; \backslash doteq\; k/n\; \backslash leq\; 1/2$.see e.g. {{harvtxt|Ash|1990|p=121}} or {{harvtxt|Flum|Grohe|2006|p=427}}.

The infinite product formula for the gamma function also gives an expression for binomial coefficients

$$(-1)^k\; \{z\; \backslash choose\; k\}=\; \{-z+k-1\; \backslash choose\; k\}\; =\; \backslash frac\{1\}\{\backslash Gamma(-z)\}\; \backslash frac\{1\}\{(k+1)^\{z+1\}\}\; \backslash prod\_\{j=k+1\}\; \backslash frac\{\backslash left(1+\backslash frac\{1\}\{j\}\backslash right)^\{-z-1\}\}\{1-\backslash frac\{z+1\}\{j\}\}$$

which yields the asymptotic formulas

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

as $k\; \backslash to\; \backslash infty$.

This asymptotic behaviour is contained in the approximation

$$\{z+k\; \backslash choose\; k\}\backslash approx\; \backslash frac\{e^\{z(H\_k-\backslash gamma)\}\}\{\backslash Gamma(z+1)\}$$

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

Further, the asymptotic formula

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

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

{{main|Multinomial theorem}}

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

: $\{n\backslash choose\; k\_1,k\_2,\backslash ldots,k\_r\}\; =\backslash frac\{n!\}\{k\_1!k\_2!\backslash cdots\; k\_r!\}$

where

: $\backslash sum\_\{i=1\}^rk\_i=n.$

While the binomial coefficients represent the coefficients of {{math|(x + y)ⁿ}}, the multinomial coefficients

represent the coefficients of the polynomial

: $(x\_1\; +\; x\_2\; +\; \backslash cdots\; +\; x\_r)^n.$

The case r = 2 gives binomial coefficients:

: $\{n\backslash choose\; k\_1,k\_2\}=\{n\backslash choose\; k\_1,\; n-k\_1\}=\{n\backslash choose\; k\_1\}=\; \{n\backslash choose\; k\_2\}.$

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:

: $\{n\backslash choose\; k\_1,k\_2,\backslash ldots,k\_r\}\; =\{n-1\backslash choose\; k\_1-1,k\_2,\backslash ldots,k\_r\}+\{n-1\backslash choose\; k\_1,k\_2-1,\backslash ldots,k\_r\}+\backslash ldots+\{n-1\backslash choose\; k\_1,k\_2,\backslash ldots,k\_r-1\}$

and symmetry:

: $\{n\backslash choose\; k\_1,k\_2,\backslash ldots,k\_r\}\; =\{n\backslash choose\; k\_\{\backslash sigma\_1\},k\_\{\backslash sigma\_2\},\backslash ldots,k\_\{\backslash sigma\_r\}\}$

where $(\backslash sigma\_i)$ is a permutation of (1, 2, ..., r).

Using Stirling numbers of the first kind the series expansion around any arbitrarily chosen point $z\_0$ is

:

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

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

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

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

: $\backslash sqrt\{1+x\}=\backslash sum\_\{k\backslash geq\; 0\}\{\backslash binom\{1/2\}\{k\}\}x^k.$

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

: $\{z\; \backslash choose\; m\}\; \{z\backslash choose\; n\}\; =\; \backslash sum\_\{k=0\}^\{\backslash min(m,n)\}\; \{m\; +\; n\; -\; k\; \backslash choose\; k,\; m\; -\; k,\; n\; -\; k\}\; \{z\; \backslash choose\; m\; +\; n\; -\; k\},$

where the connection coefficients are multinomial coefficients. In terms of labelled combinatorial objects, the connection coefficients represent the number of ways to assign {{math|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 {{math|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:

: $\backslash prod\_\{k=0\}^\{n\}\backslash binom\{n\}\{k\}=\backslash prod\_\{k=1\}^\{n\}k^\{2k-n-1\}.$

The partial fraction decomposition of the reciprocal is given by

: $\backslash frac\{1\}\{\{z\; \backslash choose\; n\}\}=\; \backslash sum\_\{i=0\}^\{n-1\}\; (-1)^\{n-1-i\}\; \{n\; \backslash choose\; i\}\; \backslash frac\{n-i\}\{z-i\},$

\qquad \frac{1}{{z+n \choose n}}= \sum_{i=1}^n (-1)^{i-1} {n \choose i} \frac{i}{z+i}.

{{main|Binomial series}}

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

: $(1+z)^\{\backslash alpha\}\; =\; \backslash sum\_\{n=0\}^\{\backslash infty\}\{\backslash alpha\backslash choose\; n\}z^n\; =\; 1+\{\backslash alpha\backslash choose1\}z+\{\backslash alpha\backslash choose\; 2\}z^2+\backslash cdots.$

The identity can be obtained by showing that both sides satisfy the differential equation {{math|1=(1 + z) f'(z) = α f(z)}}.

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

: $\backslash frac\{1\}\{(1-z)^\{\backslash alpha+1\}\}\; =\; \backslash sum\_\{n=0\}^\{\backslash infty\}\{n+\backslash alpha\; \backslash choose\; n\}z^n$

where the identity

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

is applied.

{{main|Multichoose#Counting multisets}}

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;

{{citation

| last = Munarini | first = Emanuele

| doi = 10.2298/AADM110609014M

| issue = 2

| journal = Applicable Analysis and Discrete Mathematics

| mr = 2867317

| pages = 176–200

| title = Riordan matrices and sums of harmonic numbers

| volume = 5

| year = 2011| url = http://www.doiserbia.nb.rs/img/doi/1452-8630/2011/1452-86301100014M.pdf

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

To avoid ambiguity and confusion with n's main denotation in this article,
let {{math|1=f = n = r + (k − 1)}} and {{math|1=r = f − (k − 1)}}.

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

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

One possible alternative characterization of this identity is as follows:

We may define the falling factorial as

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

and the corresponding rising factorial as

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

so, for example,

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

Then the binomial coefficients may be written as

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

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

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

{{Pascal_triangle_extended.svg}}

For any n,

:

&=(-1)^k\;\frac{n\cdot(n+1)\cdot(n+2)\cdots (n + k - 1)}{k!}\\

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

&=(-1)^k\left(\!\!\binom{n}{k}\!\!\right)\;.\end{align}

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

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

:

{-10\cdot-9\cdot-8\cdot-7\cdot-6\cdot-5\cdot-4}

{1\cdot2\cdot3\cdot4\cdot5\cdot6\cdot7}\\

&=(-1)^7\;\frac{4\cdot5\cdot6\cdot7\cdot8\cdot9\cdot10}

{1\cdot2\cdot3\cdot4\cdot5\cdot6\cdot7}\\

&=\left(\!\!\binom{-7}{7}\!\!\right)\left(\!\!\binom{4}{7}\!\!\right)=\binom{-1}{7}\binom{10}{7}.\end{align}

The binomial coefficient is generalized to two real or complex valued arguments using the gamma function or beta function via

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

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

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

moreover,

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

The resulting function has been little-studied, apparently first being graphed in {{Harv|Fowler|1996}}. Notably, many binomial identities fail: $\backslash binom\{n\; \}\{\; m\}\; =\; \backslash binom\{n\; \}\{\; n-m\}$ but $\backslash binom\{-n\}\{m\}\; \backslash neq\; \backslash binom\{-n\}\{-n-m\}$ 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 $0\; \backslash leq\; y\; \backslash leq\; x$ it is a smoothly interpolated form of the usual binomial, with a ridge ("Pascal's ridge").
• in the octant $0\; \backslash leq\; x\; \backslash leq\; y$ and in the quadrant $x\; \backslash geq\; 0,\; y\; \backslash leq\; 0$ the function is close to zero.
• in the quadrant $x\; \backslash leq\; 0,\; y\; \backslash geq\; 0$ 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.

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

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

: $\{\backslash alpha\; \backslash choose\; \backslash beta\}\; =\; \backslash left|\; \backslash left\backslash \{\; B\; \backslash subseteq\; A\; :\; \backslash left|B\backslash right|\; =\; \backslash beta\; \backslash right\backslash \}\; \backslash right|$

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

{{div col|colwidth=25em}}

• Binomial transform
• Delannoy number
• Eulerian number
• Hypergeometric function
• List of factorial and binomial topics
• Macaulay representation of an integer
• Motzkin number
• Multiplicities of entries in Pascal's triangle
• Narayana number
• Star of David theorem
• Sun's curious identity
• Table of Newtonian series
• Trinomial expansion

{{div col end}}

{{reflist|25em}}

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{{refend}}

• {{springer|title=Binomial coefficients|id=p/b016410}}
• {{cite journal |author=Andrew Granville |url=http://www.cecm.sfu.ca/organics/papers/granville/Binomial/toppage.html |title=Arithmetic Properties of Binomial Coefficients I. Binomial coefficients modulo prime powers |journal=CMS Conf. Proc |volume=20 |pages=151–162 |year=1997 |author-link=Andrew Granville |access-date=2013-09-03 |archive-url=https://web.archive.org/web/20150923201436/http://www.cecm.sfu.ca/organics/papers/granville/Binomial/toppage.html |archive-date=2015-09-23 |url-status=dead }}

{{PlanetMath attribution

| urlname = Binomialcoefficient | title = Binomial Coefficient

| urlname2 = UpperandLowerBoundstoBinomialCoefficient | title2 = Upper and lower bounds to binomial coefficient

| urlname3 = Nchooserisaninteger | title3 = Binomial coefficient is an integer

| urlname4 = Generalizedbinomialcoefficients | title4 = Generalized binomial coefficients

}}

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