Pascal's triangle: Difference between revisions

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== Combinations ==

[[File:Pascals_triangle_30_lines.png|thumb|upright=1.25|Expansion of the first 30 rows of Pascal's triangle]]

A second useful application of Pascal's triangle is in the calculation of [[combination]]s. The number of combinations of <math>n</math> items taken <math>k</math> at a time, i.e. the number of subsets of <math>k</math> elements from among <math>n</math> elements, can be found by the equation

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| last = Kallós | first = Gábor

| issue = 1

| journal = Annales Mathématiques Blaise Pascal

| pages = 1–15

| title = A generalization of Pascal's triangle using powers of base numbers

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| doi = 10.5802/ambp.211

| url = https://ambp.centre-mersenne.org/item/10.5802/ambp.211.pdf

}}.</ref> as demonstrated [[#Binomial expansions|above]]. Thus, when the entries of the row are concatenated and read in radix <math>a</math> they form the numerical equivalent of <math>(a + 1)^{n} = 11^{n}_{a}</math>. If <math>c = a + 1</math> for <math>c < 0</math>, then the theorem [[Negative base|holds]] for <math>a =\bmod 2c</math>, with <math>a</math> congruent to <math>\{c - 1, -(c + 1)\} \;\mathrm{mod}\; 2c</math>, and with odd values of <math>n</math> [[Negative number#Multiplication|yielding]] negative row products.<ref>{{cite book

| display-authors = etal

| last = Hilton | first = P.