PascalBinomial[n,m]
gives the binomial coefficient
that preserves Pascal's identity.
PascalBinomial
PascalBinomial[n,m]
gives the binomial coefficient
that preserves Pascal's identity.
Details
- Integer mathematical function, suitable for both symbolic and numerical manipulation.
- PascalBinomial is also known as combinations and as choose function.
- PascalBinomial gives the asymmetric coefficients that preserve Pascal's identity for all integer values. Use Binomial for coefficients that are symmetric for negative integer
. PascalBinomial and Binomial agree except for negative integer
. - In general,
is defined by
or suitable limits of this. - When
is a negative integer,
. » - The particular limit chosen preserves Pascal's identity
for all complex
and
. » - The symmetry rule
is satisfied for all
and most
but violated for negative integer
. » - For integer arguments, PascalBinomial automatically evaluates to exact values.
- PascalBinomial is automatically evaluated symbolically for simple cases; FunctionExpand gives results for other cases. »
- PascalBinomial can be evaluated to arbitrary numerical precision.
- PascalBinomial automatically threads over lists.
- PascalBinomial can be used with Interval and CenteredInterval objects. »
Examples
open all close allBasic Examples (5)
PascalBinomial[10, 3]Binomial[n, 3]Column[Table[PascalBinomial[n, k], {n, 0, 5}, {k, 0, n}], Center]Plot over a subset of the reals as a function of its first parameter:
Plot[PascalBinomial[x, 3], {x, -2, 4}]Plot over a subset of the reals as a function of its second parameter:
Plot[PascalBinomial[3, x], {x, -2, 4}]Plot over a subset of the complexes:
ComplexPlot3D[PascalBinomial[z, 1 / 5], {z, -2 - 2I, 2 + 2I}, PlotLegends -> Automatic]Scope (36)
Numerical Evaluation (7)
PascalBinomial[8, 4]PascalBinomial[8.5, -4.2]Evaluate for half-integer arguments:
PascalBinomial[9 / 2, 7 / 2]N[PascalBinomial[7 / 3, 1 / 5], 20]The precision of the output tracks the precision of the input:
PascalBinomial[8.2211111111115000000, 4]PascalBinomial[1 + I, 5]PascalBinomial[2. + I, 7 - 3I]Evaluate efficiently at high precision:
PascalBinomial[17 / 2, -4.2`100]//TimingPascalBinomial[-88889 + (1/2), 4.2`1000];//TimingCompute worst-case guaranteed intervals using Interval and CenteredInterval objects:
PascalBinomial[1 / 2, Interval[{0.5, 0.6}]]PascalBinomial[1, CenteredInterval[1 / 2, 1 / 100]]Or compute average-case statistical intervals using Around:
PascalBinomial[Around[2, 0.01], 2]Compute the elementwise values of an array:
PascalBinomial[ {{5π / 6, 0}, {3π / 2, -π / 2}}, 2]Or compute the matrix PascalBinomial function using MatrixFunction:
MatrixFunction[PascalBinomial[#, 2]&, {{5π / 6, 0}, {3π / 2, -π / 2}}]Specific Values (5)
Values of PascalBinomial at particular points:
Table[PascalBinomial[n, 2], {n, 1, 5}]PascalBinomial for symbolic n:
PascalBinomial[n, 4]PascalBinomial[n, 0]FunctionExpand[PascalBinomial[0, m]]Note that this is zero on all integers away from
:
Simplify[%, m∈ℤ && m != 0]PascalBinomial[n,m] is zero whenever n and m are both negative integers:
FunctionExpand[PascalBinomial[n, m], n∈NegativeIntegers && m∈NegativeIntegers]Find a value of n for which PascalBinomial[n,2]=15:
nval = n /. FindRoot[PascalBinomial[ n , 2] == 15, {n, 5}]Plot[PascalBinomial[n, 2], {n, 0, 10}, Epilog -> Style[Point[{nval, PascalBinomial[nval, 2]}], PointSize[Large], StandardMagenta]]Visualization (3)
Plot the PascalBinomial as a function of its parameter n:
Plot[{PascalBinomial[n, 1], PascalBinomial[n, 2], PascalBinomial[n, 3], PascalBinomial[n, 4]}, {n, -2, 4}]Plot the PascalBinomial as a function of its parameter
:
Plot[{PascalBinomial[1, m], PascalBinomial[2, m], PascalBinomial[3, m], PascalBinomial[4, m]}, {m, -1, 5}]ComplexContourPlot[Re[PascalBinomial[z, 5]], {z, -4 - 4I, 8 + 4I}, Contours -> 24]ComplexContourPlot[Im[PascalBinomial[z, 5]], {z, -4 - 4I, 8 + 4I}, Contours -> 24]Function Properties (12)
Real domain of PascalBinomial as a function of its parameter n:
FunctionDomain[PascalBinomial[n, 2], n]Real domain of PascalBinomial as a function of its parameter m:
FunctionDomain[PascalBinomial[2, m], m]FunctionDomain[PascalBinomial[z, 2], z, Complexes]FunctionDomain[PascalBinomial[2, z], z, Complexes]Function range of PascalBinomial:
FunctionRange[PascalBinomial[x, 2], x, y]PascalBinomial has the mirror property
:
FullSimplify[PascalBinomial[Conjugate[z], 2] == Conjugate[PascalBinomial[z, 2]]]Compute sums involving PascalBinomial:
Sum[PascalBinomial[n, k], {k, 0, n}]Sum[PascalBinomial[n, k] ^ 2, {k, 0, n}]When
is positive,
is an analytic function of both variables:
FunctionAnalytic[{PascalBinomial[x, y], x > 0}, {x, y}]This is not true for negative
:
FunctionAnalytic[{PascalBinomial[x, y], x < 0}, {x, y}]
is neither non-decreasing nor non-increasing:
FunctionMonotonicity[PascalBinomial[x, 7], x]FunctionInjective[PascalBinomial[x, 7], x]Plot[{PascalBinomial[x, 7], .01}, {x, -2, 3}]FunctionSurjective[PascalBinomial[x, 7], x]Plot[{PascalBinomial[x, 7], 750}, {x, -10, 15}]PascalBinomial is neither non-negative nor non-positive:
FunctionSign[PascalBinomial[x, 1], x]FunctionSign[PascalBinomial[1, x], x]
has singularities and discontinuities where
is a negative integer:
FunctionSingularities[PascalBinomial[x, y], x]FunctionDiscontinuities[PascalBinomial[x, y], x]
is neither convex nor concave:
FunctionConvexity[PascalBinomial[x, 7], x]TraditionalForm formatting:
PascalBinomial[n, k]//TraditionalFormDifferentiation (3)
First derivative with respect to
:
D[PascalBinomial[n, m], m]Higher derivatives with respect to
:
Table[D[PascalBinomial[n, m], {m, k}], {k, 1, 3}]//FullSimplifyPlot the higher derivatives with respect to
for
:
Plot[Evaluate[% /. n -> 2], {m, -5, 5}, PlotLegends -> {"First Derivative", "Second Derivative", "Third Derivative"}]First derivative with respect to
:
D[PascalBinomial[n, m], n]Series Expansions (4)
Find the Taylor expansion using Series:
Series[PascalBinomial[n, x], {x, 0, 3}]//Normal//FullSimplifyPlots of the first three approximations around
:
terms = Normal@Table[Series[PascalBinomial[2, x], {x, 0, m}], {m, 1, 5, 2}];
Plot[{PascalBinomial[2, x], terms}, {x, -5, 5}]Find the series expansion at Infinity:
FullSimplify[Series[PascalBinomial[n, x], {x, Infinity, 1}]//Normal, x > n]Find series expansion for an arbitrary symbolic direction
:
Series[PascalBinomial[2, x], {x, DirectedInfinity[z], 1}, Assumptions -> x > 0 ]//Normal// FullSimplifyTaylor expansion at a generic point:
Series[PascalBinomial[n, x], {x, x0, 2}]// Normal//FullSimplifyFunction Identities and Simplifications (2)
PascalBinomial[n, k] == (k + 1/n - k) PascalBinomial[n, k + 1]//FullSimplifyPascalBinomial[n, k] == (n/n - k) PascalBinomial[n - 1, k]//FullSimplifyPascalBinomial[n, k] == (n - k + 1/n + 1) PascalBinomial[n + 1, k]//FullSimplifyApplications (9)
There are
ways to choose
elements without replacements from a set of
elements:
With[{n = 5, m = 2}, PascalBinomial[n, m]]Check with direct enumeration:
With[{n = 5, m = 2}, Subsets[Range[n], {m}]]Length[%]There are
ways to choose
elements with replacement from a set of
elements:
With[{n = 5, m = 2}, PascalBinomial[n + m - 1, m]]Check with direct enumeration:
With[{n = 5, m = 2}, DeleteDuplicates[Sort /@ Tuples[Range[n], {m}]]]Length[%]There are
ways to arrange
indistinguishable objects of one kind and
indistinguishable objects of another kind:
With[{n = 5, m = 2}, PascalBinomial[n + m, m]]With[{n = 5, m = 2}, Permutations[Join[ConstantArray[0, n], ConstantArray[1, m]]]]Length[%]Illustrate the binomial theorem:
Sum[PascalBinomial[n, k]x ^ k, {k, 0, n}]Expand[(1 + x) ^ 10]Series[(1 + x) ^ (1 / 3), {x, 0, 5}]Sum[PascalBinomial[1 / 3, k]x ^ k, {k, 0, 5}]ArrayPlot[Table[Mod[PascalBinomial[i, j], 2], {i, 50}, {j, i}], PlotTheme -> "Scientific"]Plot PascalBinomial in the arguments' plane:
Plot3D[PascalBinomial[n, k], {n, -10, 10}, {k, -10, 10}]Plot the logarithm of the number of ways to pick
elements out of
:
Plot3D[Log[PascalBinomial[n, k]], {n, 1, 12}, {k, 1, 12}, RegionFunction -> (#2 <= #1&)]Compute higher derivatives of a product of two functions:
Sum[PascalBinomial[4, k] D[f[z], {z, k}]D[g[z], {z, 4 - k}], {k, 0, 4}]D[f[z] g[z], {z, 4}]Properties & Relations (8)
On the integers, PascalBinomial[n,m] equals
:
And@@Flatten@Table[PascalBinomial[n, m] == Underscript[, νUnderscript[ -> , ℂ]n](Underscript[, μUnderscript[ -> , ℂ]m](Gamma[ν + 1]/Gamma[μ + 1] Gamma[-μ + ν + 1])), {n, -5, 5}, {m, -5, 5}]This can be expressed as
for
and
otherwise:
And@@Flatten@Table[PascalBinomial[``n``, ``m``] == Piecewise[{{((-1)^m*Pochhammer[-n, m])/m!, m >= 0}, {0, m <= n}}], {n, 5, -5, -1}, {m, 5, -5, -1}]An alternative formula on the integers:
Assuming[{n, m}∈ℤ, FullSimplify@FunctionExpand[PascalBinomial[n, m]]]Pascal's identity is satisfied everywhere:
FullSimplify[PascalBinomial[n, m] == PascalBinomial[n - 1, m] + PascalBinomial[n - 1, m - 1]]In particular, it is true at the origin:
FullSimplify[PascalBinomial[n, m] == PascalBinomial[n - 1, m] + PascalBinomial[n - 1, m - 1], n == 0 && m == 0]The symmetry rule
may not hold for negative integer
:
FullSimplify[PascalBinomial[n, m] == PascalBinomial[n, n - m], n∈NegativeIntegers && m∈ℤ]It may hold for some values but will generally be violated for positive integer
:
PascalBinomial[n, m] == PascalBinomial[n, n - m] /. {{n -> -3, m -> 1}, {n -> -3, m -> -1}}Binomial satisfies the symmetry rule everywhere:
FullSimplify[Binomial[n, m] == Binomial[n, n - m]]PascalBinomial performs simple evaluations for symbolic arguments:
PascalBinomial[n, 2]PascalBinomial typically does not evaluate when both arguments are symbolic:
PascalBinomial[n + 1, n]Use FunctionExpand with conditions to achieve appropriate simplifications:
FunctionExpand[PascalBinomial[n + 1, n], n∈Integers]PascalBinomial[n,m] is 0 whenever n and m are both negative integers:
FunctionExpand[PascalBinomial[n, m], n∈NegativeIntegers && m∈NegativeIntegers]Use FullSimplify to simplify expressions involving binomial coefficients:
FullSimplify[PascalBinomial[n + 1, k + 1] / PascalBinomial[n, k]]Use FunctionExpand to expand into Gamma functions:
FunctionExpand[PascalBinomial[n, k]]Sums involving PascalBinomial:
Underoverscript[∑, k = 0, n](-1)^kPascalBinomial[n, k]Underoverscript[∑, k = 0, ∞]PascalBinomial[n, k] kUnderoverscript[∑, k = 1, ∞](1/PascalBinomial[2 k, k])Neat Examples (7)
Construct a graphical version of Pascal's triangle:
With[{n = 8},
Graph[Flatten[Table[{b[i, j] <-> b[i + 1, j], b[i, j] <-> b[i + 1, j + 1], b[i + 1, j] <-> b[i + 1, j + 1]}, {i, 0, n - 1}, {j, 0, i}]], VertexCoordinates -> {b[i_, j_] :> {(n + 1/2) - i / 2 + j, -i}}, VertexLabels -> {b[i_, j_] :> Placed[PascalBinomial[i, j], Center]}, IconizedObject[«Style options»]]]Extend the triangle to negative integers; unlabeled points indicate a zero value:
With[{n = 6}, Graph[Flatten[Table[{b[i, j] <-> b[i + 1, j], b[i, j] <-> b[i + 1, j + 1], b[i + 1, j] <-> b[i + 1, j + 1]}, {i, n - 1, -n - 1, -1}, {j, -n + i UnitStep[i] - UnitStep[-1 - i], n + i UnitStep[-i] + UnitStep[-1 - i]}]], VertexCoordinates -> {b[i_, j_] :> {(n + 1/2) - (i/2) + j, -i}}, IconizedObject[«Style options»]]]Binomial, by contrast, defines the top-left sector by reflecting the top-right sector:
With[{n = 6}, Graph[Flatten[Table[{b[i, j] <-> b[i + 1, j], b[i, j] <-> b[i + 1, j + 1], b[i + 1, j] <-> b[i + 1, j + 1]}, {i, n - 1, -n - 1, -1}, {j, -n + i UnitStep[i] - UnitStep[-1 - i], n + i UnitStep[-i] + UnitStep[-1 - i]}]], VertexCoordinates -> {b[i_, j_] :> {(n + 1/2) - (i/2) + j, -i}}, IconizedObject[«Style options»]]]ArrayPlot[Table[Mod[PascalBinomial[i, j], Pi], {i, 50}, {j, i}], PlotTheme -> "Scientific"]Closed‐form inverse of Hilbert matrices:
Block[{n = 4}, Table[(-1) ^ (i + j)(i + j - 1)PascalBinomial[n + i - 1, n - j]PascalBinomial[n + j - 1, n - i]PascalBinomial[i + j - 2, i - 1] ^ 2, {i, n}, {j, n}]]//MatrixFormInverse[%]Nested binomials over the complex plane:
DensityPlot[Arg[Nest[PascalBinomial[#, # / 2]&, x + I y, 3]], {x, -1, -0.9}, {y, -0.06, 0.06}, Exclusions -> {}, MaxRecursion -> 4, PlotPoints -> 50]Plot PascalBinomial at infinity:
DensityPlot[Arg[PascalBinomial[(x + I y), 1 / (x + I y)]], {x, -1 / 2, 1 / 2}, {y, -1 / 2, 1 / 2}, Exclusions -> {}, MaxRecursion -> 4, PlotPoints -> 50]Plot PascalBinomial for complex arguments:
DensityPlot[Arg[PascalBinomial[4.1 Exp[I ϕ1], 2.2 Exp[I ϕ2]]], {ϕ1, -Pi, Pi}, {ϕ2, -Pi, Pi}, Exclusions -> {}, MaxRecursion -> 4, PlotPoints -> 50]Plot PascalBinomial at Gaussian integers:
Block[{$MaxExtraPrecision = 1000}, ArrayPlot[Table[Mod[Im@Round[PascalBinomial[x + I y, x ]], 2], {x, 0, 50}, {y, -25, 25}], PlotTheme -> "Scientific"]]History
Text
Wolfram Research (2024), PascalBinomial, Wolfram Language function, https://reference.wolfram.com/language/ref/PascalBinomial.html.
CMS
Wolfram Language. 2024. "PascalBinomial." Wolfram Language & System Documentation Center. Wolfram Research. https://reference.wolfram.com/language/ref/PascalBinomial.html.
APA
Wolfram Language. (2024). PascalBinomial. Wolfram Language & System Documentation Center. Retrieved from https://reference.wolfram.com/language/ref/PascalBinomial.html
BibTeX
@misc{reference.wolfram_2026_pascalbinomial, author="Wolfram Research", title="{PascalBinomial}", year="2024", howpublished="\url{https://reference.wolfram.com/language/ref/PascalBinomial.html}", note=[Accessed: 12-June-2026]}
BibLaTeX
@online{reference.wolfram_2026_pascalbinomial, organization={Wolfram Research}, title={PascalBinomial}, year={2024}, url={https://reference.wolfram.com/language/ref/PascalBinomial.html}, note=[Accessed: 12-June-2026]}