View source for Pascal's triangle

{{short description|Triangular array of the binomial coefficients in mathematics}}
{{Image frame|width=230|caption=A diagram showing the first eight rows of Pascal's triangle.
|content=
<math>
\begin{array}{c}
 1 \\
 1 \quad 1 \\
 1 \quad 2 \quad 1 \\
 1 \quad 3 \quad 3 \quad 1 \\
 1 \quad 4 \quad 6 \quad 4 \quad 1 \\
 1 \quad 5 \quad 10 \quad 10 \quad 5 \quad 1 \\
 1 \quad 6 \quad 15 \quad 20 \quad 15 \quad 6 \quad 1 \\
 1 \quad 7 \quad 21 \quad 35 \quad 35 \quad 21 \quad 7 \quad 1
\end{array}
</math>
}}In [[mathematics]], '''Pascal's triangle''' is an infinite [[triangular array]] of the [[binomial coefficient]]s which play a crucial role in probability theory, [[combinatorics]], and algebra. In much of the [[Western world]], it is named after the French mathematician [[Blaise Pascal]], although other [[mathematician]]s studied it centuries before him in Persia,<ref name=":0" /> India,<ref>Maurice Winternitz, ''History of Indian Literature'', Vol. III</ref> China, Germany, and Italy.<ref name="roots">{{cite book |author=Peter Fox |title=Cambridge University Library: the great collections |url=https://books.google.com/books?id=xxlgKP5thL8C&pg=PA13 |year=1998 |publisher=Cambridge University Press |isbn=978-0-521-62647-7 |page=13}}</ref>

The rows of Pascal's triangle are conventionally enumerated starting with row <math>n = 0</math> at the top (the 0th row). The entries in each row are numbered from the left beginning with <math>k = 0</math> and are usually staggered relative to the numbers in the adjacent rows. The triangle may be constructed in the following manner: In row 0 (the topmost row), there is a unique nonzero entry 1. Each entry of each subsequent row is constructed by adding the number above and to the left with the number above and to the right, treating blank entries as 0. For example, the initial number of row 1 (or any other row) is 1 (the sum of 0 and 1), whereas the numbers 1 and 3 in row 3 are added to produce the number 4 in row 4.

== Formula ==
[[File:PascalTriangleAnimated2.gif|thumb|upright=1|In Pascal's triangle, each number is the sum of the two numbers directly above it.]]In the <math>n</math>th row of Pascal's triangle, the <math>k</math>th entry is denoted <math>\tbinom nk</math>, pronounced "{{mvar|n}} choose {{mvar|k}}". For example, the topmost entry is <math>\tbinom 00 = 1</math>. With this notation, the construction of the previous paragraph may be written as
<math display="block"> {n \choose k} = {n-1 \choose k-1} + {n-1 \choose k}</math>
for any positive integer <math>n</math> and any integer <math>0 \le k \le n</math>.<ref>The binomial coefficient <math>\scriptstyle {n \choose k}</math> is conventionally set to zero if ''k'' is either less than zero or greater than ''n''.</ref>  This recurrence for the binomial coefficients is known as [[Pascal's rule]].

== History ==
[[File:Yanghui triangle.gif|thumb|right|upright=1|[[Yang Hui]]'s triangle, as depicted by the Chinese using [[Counting rods|rod numerals]], appears in [[Jade Mirror of the Four Unknowns]], a mathematical work by [[Zhu Shijie]], dated 1303.]]
[[File:TrianguloPascal.jpg|thumb|right|upright=1.25|[[Blaise Pascal|Pascal]]'s version of the triangle]]

The pattern of numbers that forms Pascal's triangle was known well before Pascal's time. The Persian mathematician [[Al-Karaji]] (953–1029) wrote a now-lost book which contained the first description of Pascal's triangle.<ref>{{Cite book|url=https://books.google.com/books?id=kt9DIY1g9HYC&q=al+karaji+pascal%27s+triangle&pg=PA132|title=Encyclopaedia of the History of Science, Technology, and Medicine in Non-Western Cultures|last=Selin|first=Helaine|date=2008-03-12|publisher=Springer Science & Business Media|isbn=9781402045592|language=en|page=132|bibcode=2008ehst.book.....S|quote=Other, lost works of al-Karaji's are known to have dealt with inderterminate algebra, arithmetic, inheritance algebra, and the construction of buildings. Another contained the first known explanation of the arithmetical (Pascal's) triangle; the passage in question survived through al-Sama'wal's Bahir (twelfth century) which heavily drew from the Badi.}}</ref><ref>{{Cite book |last=Rashed |first=R. |url=https://books.google.com/books?id=vSkClSvU_9AC&pg=PA62 |title=The Development of Arabic Mathematics: Between Arithmetic and Algebra |date=1994-06-30 |publisher=Springer Science & Business Media |isbn=978-0-7923-2565-9 |pages=63 |language=en}}</ref><ref>{{Cite book|url=https://books.google.com/books?id=kAjABAAAQBAJ&q=al+karaji+binomial+theorem&pg=PA54|title=From Alexandria, Through Baghdad: Surveys and Studies in the Ancient Greek and Medieval Islamic Mathematical Sciences in Honor of J.L. Berggren|last1=Sidoli|first1=Nathan|last2=Brummelen|first2=Glen Van|date=2013-10-30|publisher=Springer Science & Business Media|isbn=9783642367366|language=en|page=54|quote=However, the use of binomial coefficients by Islamic mathematicians of the eleventh century, in a context which had deep roots in Islamic mathematics, suggests strongly the table was a local discovery - most probably of al-Karaji."}}</ref> In India, the ''[[Chandaḥśāstra]]'' by the Indian poet and mathematician [[Piṅgala]] (3rd or 2nd century BC) somewhat cryptically describes a method of arranging two types of syllables to form [[metre (poetry)|metre]]s of various lengths and counting them; as interpreted and elaborated by Piṅgala's 10th-century commentator [[Halāyudha]] his "method of pyramidal expansion" (''meru-prastāra'') for counting metres is equivalent to [[Pascal's triangle]].<ref>{{cite journal |last=Alsdorf |first=Ludwig |year=1991 |orig-year=1933 |title=The Pratyayas: Indian Contribution to Combinatorics |journal=Indian Journal of History of Science |volume=26 |number=1 |pages=17–61 |url=https://insa.nic.in/(S(f0a4mvblfb5vfmialsdau2an))/writereaddata/UpLoadedFiles/IJHS/Vol26_1_3_SRSarma.pdf}} Translated by S. R. Sarma from {{cite journal |last=Alsdorf |first=Ludwig |display-authors=0 |title={{mvar|π}} Die Pratyayas. Ein Beitrag zur indischen Mathematik |journal=Zeitschrift fur Indologie und Iranistik |volume=9 |year=1933 |pages=97–157 }} {{pb}} {{cite journal |last=Bag |first=Amulya Kumar |title=Binomial theorem in ancient India |journal=Indian Journal of History of Science |volume=1 |number=1 |year=1966 |pages=68–74 |url=http://repository.ias.ac.in/70374/1/10-pub.pdf }} {{pb}} Tertiary sources: {{pb}} {{cite book |title=A Concise History Of Science In India |year=1971 |publisher=Indian National Science Academy |editor-last=Bose |editor-first=D. M. |editor-link=Debendra Mohan Bose |last=Sen |first=Samarendra Nath |chapter=Mathematics |chapter-url=https://archive.org/details/in.ernet.dli.2015.502083/page/n178 |at=Ch. 3, {{pgs|136–212}}, esp. "Permutations, Combinations and Pascal Triangle", {{pgs|156–157}} }} {{pb}} {{cite journal |title=The Binomial Coefficient Function |last=Fowler |first=David H. |author-link= David Fowler (mathematician) |journal=The American Mathematical Monthly |year=1996 |volume=103 |number=1 |pages=1–17, esp. §4 "A Historical Note", {{pgs|10–17}} |doi=10.2307/2975209 |jstor=2975209 }}</ref> It was later repeated by [[Omar Khayyám]] (1048–1131), another Persian mathematician; thus the triangle is also referred to as '''Khayyam's triangle''' ({{langx|fa|مثلث خیام|label=none}}) in Iran.<ref>{{cite book |author=Kennedy, E. |title=Omar Khayyam. The Mathematics Teacher 1958 |jstor=i27957284|year=1966 |publisher=National Council of Teachers of Mathematics |pages=140–142}}</ref> Several theorems related to the triangle were known, including the [[binomial theorem]]. Khayyam used a method of finding [[nth root|''n''th roots]] based on the binomial expansion, and therefore on the binomial coefficients.<ref name=":0">{{citation
 | last = Coolidge | first = J. L. | author-link = Julian Coolidge
 | journal = [[The American Mathematical Monthly]]
 | jstor = 2305028
 | mr = 0028222
 | pages = 147–157
 | title = The story of the binomial theorem
 | volume = 56
 | issue = 3 | year = 1949| doi = 10.2307/2305028 }}.</ref>

Pascal's triangle was known in China during the 11th century through the work of the Chinese mathematician [[Jia Xian]] (1010–1070). During the 13th century, [[Yang Hui]] (1238–1298) defined the triangle, and it is known as '''Yang Hui's triangle''' ({{lang-zh|s=杨辉三角|t=楊輝三角|labels=no}}) in China.<ref>Weisstein, Eric W. (2003). ''CRC concise encyclopedia of mathematics'', p. 2169. {{isbn|978-1-58488-347-0}}.</ref>

In Europe, Pascal's triangle appeared for the first time in the ''Arithmetic'' of [[Jordanus de Nemore]] (13th century).<ref>{{cite journal |last1=Hughes |first1=Barnabas |title=The arithmetical triangle of Jordanus de Nemore |journal=Historia Mathematica |date=1 August 1989 |volume=16 |issue=3 |pages=213–223 |doi=10.1016/0315-0860(89)90018-9 |doi-access=free }}</ref>
The binomial coefficients were calculated by [[Gersonides]] during the early 14th century, using the multiplicative formula for them.<ref name="ed-cam">{{citation|contribution=The arithmetical triangle|first=A. W. F.|last=Edwards|pages=166–180|title=Combinatorics: Ancient and Modern|publisher=Oxford University Press|year=2013|editor1-first=Robin|editor1-last=Wilson|editor2-first=John J.|editor2-last=Watkins}}.</ref> [[Petrus Apianus]] (1495–1552) published the full triangle on the [[Book frontispiece|frontispiece]] of his book on business calculations in 1527.<ref>{{citation|title=Nature of Mathematics|first=Karl J.|last=Smith|publisher=Cengage Learning|year=2010|isbn=9780538737586|page=10|url=https://books.google.com/books?id=Di0HyCgDYq8C&pg=PA10}}.</ref> [[Michael Stifel]] published a portion of the triangle (from the second to the middle column in each row) in 1544, describing it as a table of [[figurate number]]s.<ref name="ed-cam"/> In Italy, Pascal's triangle is referred to as '''Tartaglia's triangle''', named for the Italian algebraist [[Nicolo Tartaglia|Tartaglia]] (1500–1577), who published six rows of the triangle in 1556.<ref name="ed-cam"/> [[Gerolamo Cardano]] also published the triangle as well as the additive and multiplicative rules for constructing it in 1570.<ref name="ed-cam"/>

Pascal's {{lang|fr|Traité du triangle arithmétique}} (''Treatise on Arithmetical Triangle'') was published posthumously in 1665.<ref>{{Cite book |last=Pascal |first=Blaise (1623-1662) Auteur du texte |url=https://gallica.bnf.fr/ark:/12148/btv1b86262012/f1.image |title=Traité du triangle arithmétique , avec quelques autres petits traitez sur la mesme matière. Par Monsieur Pascal |date=1665 |language=EN}}</ref>   In this, Pascal collected several results then known about the triangle, and employed them to solve problems in [[probability theory]]. The triangle was later named for Pascal by [[Pierre Raymond de Montmort]] (1708) who called it {{lang|fr|table de M. Pascal pour les combinaisons}} (French: Mr. Pascal's table for combinations) and [[Abraham de Moivre]] (1730) who called it {{lang|la|Triangulum Arithmeticum PASCALIANUM}} (Latin: Pascal's Arithmetic Triangle), which became the basis of the modern Western name.<ref>{{Cite journal | doi = 10.2307/2975209 | title = The Binomial Coefficient Function | first =  David | last = Fowler | author-link = David Fowler (mathematician) | journal = [[The American Mathematical Monthly]] | volume = 103 | issue = 1 |date=January 1996 | pages = 1–17 | jstor = 2975209 }} See in particular p. 11.</ref>

== Binomial expansions ==
[[File:Binomial theorem visualisation.svg|thumb|upright=1.25|Visualisation of binomial expansion up to the 4th power]]

Pascal's triangle determines the coefficients which arise in [[binomial expansion]]s. For example, in the expansion
<math display="block">(x + y)^2 = x^2 + 2xy + y^2 = \mathbf{1} x^2 y^0 + \mathbf{2} x^1 y^1 + \mathbf{1} x^0 y^2,</math>
the coefficients are the entries in the second row of Pascal's triangle: <math>\tbinom 20 = 1</math>, <math>\tbinom 21 = 2</math>, <math>\tbinom 22 = 1</math>.

In general, the [[binomial theorem]] states that when a [[binomial (polynomial)|binomial]] like <math>x + y</math> is raised to a positive integer power <math>n</math>, the expression expands as
<math display="block">(x + y)^n = 
\sum_{k=0}^{n} a_{k} x^{n-k} y^{k} = 
a_{0} x^n + a_{1} x^{n - 1} y + a_{2} x^{n - 2} y^{2} + \ldots + a_{n - 1} x y^{n-1} + a_{n} y^{n}, </math>
where the coefficients <math>a_{k} </math> are precisely the numbers in row <math>n </math> of Pascal's triangle:
<math display="block">a_k = {n \choose k}.</math>

The entire left diagonal of Pascal's triangle corresponds to the coefficient of <math>x^n </math> in these binomial expansions, while the next left diagonal corresponds to the coefficient of <math>x^{n-1} y </math>, and so on.

To see how the binomial theorem relates to the simple construction of Pascal's triangle, consider the problem of calculating the coefficients of the expansion of <math>(x + y)^{n + 1} </math> in terms of the corresponding coefficients of <math>(x + 1)^{n} </math>, where we set <math>y = 1 </math> for simplicity. Suppose then that
<math display="block">
(x + 1)^{n} = \sum_{k = 0}^{n} a_{k} x^{k}.
</math>
Now
<math display="block"> (x+1)^{n+1} = (x+1)(x+1)^n = x(x+1)^n + (x+1)^n = \sum_{i=0}^n a_i x^{i+1} + \sum_{k=0}^n a_k x^k.</math>

{{Image frame|width=260|caption=The first six rows of Pascal's triangle as binomial coefficients
|content=
<math>\begin{array}{c}
\dbinom{0}{0} \\
                                \dbinom{1}{0} \quad \dbinom{1}{1} \\
                        \dbinom{2}{0} \quad \dbinom{2}{1} \quad \dbinom{2}{2} \\
                \dbinom{3}{0} \quad \dbinom{3}{1} \quad \dbinom{3}{2} \quad \dbinom{3}{3} \\
        \dbinom{4}{0} \quad \dbinom{4}{1} \quad \dbinom{4}{2} \quad \dbinom{4}{3} \quad \dbinom{4}{4} \\
\dbinom{5}{0} \quad \dbinom{5}{1} \quad \dbinom{5}{2} \quad \dbinom{5}{3} \quad \dbinom{5}{4} \quad \dbinom{5}{5}
\end{array}</math>
}}

The two summations can be reindexed with <math>k=i+1 </math> and combined to yield
<math display="block">
\begin{align}
\sum_{i=0}^{n} a_{i} x^{i+1} + \sum_{k=0}^n a_k x^k &= 
\sum_{k=1}^{n+1} a_{k-1} x^{k} + \sum_{k=0}^n a_k x^k \\ [4pt] &= 
\sum_{k=1}^{n} a_{k-1} x^{k}  + a_{n}x^{n+1} + a_0x^0 + \sum_{k=1}^n a_k x^k \\[4pt] &= 
a_0x^0 + \sum_{k=1}^{n} (a_{k-1} + a_k)x^{k} + a_{n}x^{n+1} \\[4pt] &= 
x^0 + \sum_{k=1}^{n} (a_{k-1} + a_k)x^{k} + x^{n+1}.
\end{align}
</math>

Thus the extreme left and right coefficients remain as 1, and for any given <math>0 < k < n + 1 </math>, the coefficient of the <math>x^{k} </math> term in the polynomial <math>(x + 1)^{n + 1} </math> is equal to <math>a_{k-1} + a_{k} </math>, the sum of the <math>x^{k-1} </math> and <math>x^{k} </math> coefficients in the previous power <math>(x + 1)^n </math>. This is indeed the downward-addition rule for constructing Pascal's triangle.

It is not difficult to turn this argument into a [[proof (mathematics)|proof]] (by [[mathematical induction]]) of the binomial theorem. 

Since <math>(a + b)^{n} = b^{n}(\tfrac{a}{b} + 1 )^{n} </math>, the coefficients are identical in the expansion of the general case.

An interesting consequence of the binomial theorem is obtained by setting both variables <math>x = y = 1 </math>, so that
<math display="block"> \sum_{k = 0}^{n} {n \choose k} 
= {n \choose 0} + {n \choose 1} + \cdots + {n \choose n-1} + {n \choose n}
= (1+1)^n = 2^{n}. </math>

In other words, the sum of the entries in the <math>n</math>th row of Pascal's triangle is the <math>n</math>th power of&nbsp;2. This is equivalent to the statement that the number of subsets of an <math>n</math>-element set is <math>2^n</math>, as can be seen by observing that each of the <math>n</math> elements may be independently included or excluded from a given subset.

== Combinations ==
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

:<math> \mathbf{C}(n, k) = \mathbf{C}_{k}^{n}= {{}_{n}C_{k}} = {n \choose k} = \frac{n!}{k!(n-k)!}</math>.

This is equal to entry <math>k</math> in row <math>n</math> of Pascal's triangle. Rather than performing the multiplicative calculation, one can simply look up the appropriate entry in the triangle (constructed by additions). For example, suppose 3 workers need to be hired from among 7 candidates; then the number of possible hiring choices is 7 choose 3, the entry 3 in row 7 of the above table (taking into consideration the first row is the 0th row), which is <math> \tbinom{7}{3}=35 </math>.<ref>{{Cite web |url=http://5010.mathed.usu.edu/Fall2018/HWheeler/probability.html |access-date=2023-06-01 |website=5010.mathed.usu.edu|title=Pascal's Triangle in Probability}}</ref>

== Relation to binomial distribution and convolutions ==
When divided by <math> 2^n</math>, the <math> n</math>th row of Pascal's triangle becomes the [[binomial distribution]] in the symmetric case where <math> p = \tfrac{1}{2}</math>. By the [[central limit theorem]], this distribution approaches the [[normal distribution]] as <math> n</math> increases. This can also be seen by applying [[Stirling's formula]] to the factorials involved in the formula for combinations.

This is related to the operation of [[discrete convolution]] in two ways. First, polynomial multiplication corresponds exactly to discrete convolution, so that repeatedly convolving the sequence <math> \{ \ldots, 0, 0, 1, 1, 0, 0, \ldots \}</math> with itself corresponds to taking powers of <math> x + 1</math>, and hence to generating the rows of the triangle. Second, repeatedly convolving the distribution function for a [[random variable]] with itself corresponds to calculating the distribution function for a sum of ''n'' independent copies of that variable; this is exactly the situation to which the central limit theorem applies, and hence results in the normal distribution in the limit. (The operation of repeatedly taking a convolution of something with itself is called the [[convolution power]].)

== Patterns and properties ==
Pascal's triangle has many properties and contains many patterns of numbers.
[[File:Pascal's Triangle animated binary rows.gif|thumb|upright=1|Each frame represents a row in Pascal's triangle. Each column of pixels is a number in binary with the least significant bit at the bottom. Light pixels represent 1 and dark pixels 0.]]
[[File:pascal_triangle_compositions.svg|thumb|upright=1|The numbers of [[composition (combinatorics)|compositions]] of ''n''&hairsp;+1 into ''k''&hairsp;+1 ordered partitions form Pascal's triangle.]]

=== Rows ===
* The sum of the elements of a single row is twice the sum of the row preceding it. For example, row&nbsp;0 (the topmost row) has a value of 1, row&nbsp;1 has a value of 2, row&nbsp;2 has a value of 4, and so forth. This is because every item in a row produces two items in the next row: one left and one right. The sum of the elements of row&nbsp;<math> n</math> equals to <math> 2^n</math>.
*Taking the product of the elements in each row, the sequence of products {{OEIS|id=A001142}} is related to the base of the natural logarithm, ''[[E (mathematical constant)|e]]''.<ref>{{citation
 | last = Brothers | first = H. J.
 | doi = 10.4169/math.mag.85.1.51
 | journal = [[Mathematics Magazine]]
 | pages = 51
 | title = Finding e in Pascal's triangle
 | volume = 85
 | year = 2012| issue = 1
 | s2cid = 218541210
 }}.</ref><ref>{{citation
 | last = Brothers | first = H. J.
 | doi =10.1017/S0025557200004204
 | journal = [[The Mathematical Gazette]]
 | pages = 145–148
 | title = Pascal's triangle: The hidden stor-''e''
 | volume = 96
 | year = 2012| issue = 535
 | s2cid = 233356674
 }}.</ref> Specifically, define the sequence <math> s_{n}</math>  for all  <math>n \ge 0</math> as follows: <math>s_{n} = \prod_{k = 0}^{n} {n \choose k} = \prod_{k = 0}^{n} \frac{n!}{k!(n-k)!}</math> {{pb}} Then, the ratio of successive row products is <math display="block">\frac{s_{n+1}}{s_{n}} = 
\frac{ \displaystyle (n+1)!^{n+2} \prod_{k = 0}^{n + 1} \frac{1}{k!^2}}{\displaystyle n!^{n+1}\prod_{k=0}^{n}{\frac{1}{k!^2}}} = \frac{(n + 1)^n}{n!}</math> and the ratio of these ratios is <math display="block">\frac{s_{n + 1} \cdot s_{n - 1}}{s_{n}^{2}} = 
\left( \frac{n + 1}{n} \right)^n, ~ n\ge 1.</math> The right-hand side of the above equation takes the form of the limit definition of [[e (mathematical constant)|<math>e</math>]] <math display="block">e =\lim_{n \to \infty} \left( 1 + \frac{1}{n} \right)^{n}</math>.
* [[pi|<math>\pi</math>]] can be found in Pascal's triangle by use of the [[Nilakantha Somayaji|Nilakantha]] infinite series.<ref> {{citation
 | last = Foster | first = T.
 | doi = 10.5951/mathteacher.108.4.0246
 | journal = [[Mathematics Teacher]]
 | pages = 247
 | title = Nilakantha's Footprints in Pascal's Triangle
 | volume = 108
 | year = 2014}}</ref> <math display="block">\pi = 3 + \sum_{n = 1}^{\infty} (-1)^{n + 1} \frac{{2n + 1 \choose 1}}{{2n + 1 \choose 2}{2n + 2 \choose 2}}</math>
* Some of the numbers in Pascal's triangle correlate to numbers in [[Lozanić's triangle]].
* The sum of the squares of the elements of row&nbsp;{{mvar|n}} equals the middle element of row&nbsp;{{math|2''n''}}. For example, {{math|1=1<sup>2</sup> + 4<sup>2</sup> + 6<sup>2</sup> + 4<sup>2</sup> + 1<sup>2</sup> = 70}}. In general form, <math display="block">\sum_{k=0}^n {n \choose k}^2 = {2n \choose n}.</math>
* In any even row <math>n=2m</math>, the middle term minus the term two spots to the left equals a [[Catalan number]], specifically <math>C_{m-1} = \tbinom{2m}{m} - \tbinom{2m}{m-2}</math>. For example, in row&nbsp;4, which is 1, 4, 6, 4, 1, we get the 3rd Catalan number <math>C_3 = 6-1 = 5 </math>.
* In a row&nbsp;{{mvar|p}}, where {{mvar|p}} is a [[prime number]], all the terms in that row except the 1s are divisible by {{mvar|p}}. This can be proven easily, from the multiplicative formula <math>\tbinom pk = \tfrac{p!}{k!(p-k)!} </math>. Since the denominator <math>k!(p-k)!</math> can have no prime factors equal to {{mvar|p}}, so {{mvar|p}} remains in the numerator after integer division, making the entire entry a multiple of {{mvar|p}}.<!--
[[Image:Exp binomial grey.svg|thumb|upright=1.25|left|Binomial matrix as matrix exponential (illustration for 5×5 matrices). All the dots represent 0.]]  Comment: picture hidded since useless rescaling, also wrong paragraph for the pic -->
* ''Parity'': To count [[odd number|odd]] terms in row&nbsp;{{mvar|n}}, convert {{mvar|n}} to [[binary numeral system|binary]]. Let {{mvar|x}} be the number of 1s in the binary representation. Then the number of odd terms will be {{math|2<sup>''x''</sup>}}. These numbers are the values in [[Gould's sequence]].<ref>{{citation
 | last = Fine | first = N. J.
 | doi = 10.2307/2304500
 | journal = [[American Mathematical Monthly]]
 | mr = 0023257
 | pages = 589–592
 | title = Binomial coefficients modulo a prime
 | volume = 54
 | issue = 10
 | year = 1947| jstor = 2304500
 }}. See in particular Theorem 2, which gives a generalization of this fact for all prime moduli.</ref>
* Every entry in row 2<sup>''n''</sup>&nbsp;−&nbsp;1, ''n''&nbsp;≥&nbsp;0, is odd.<ref>{{citation
 | last = Hinz | first = Andreas M.
 | doi = 10.2307/2324061
 | issue = 6
 | journal = The American Mathematical Monthly
 | mr = 1166003
 | pages = 538–544
 | title = Pascal's triangle and the Tower of Hanoi
 | volume = 99
 | year = 1992| jstor = 2324061
 }}. Hinz attributes this observation to an 1891 book by [[Édouard Lucas]], ''Théorie des nombres'' (p.&nbsp;420).</ref>
*''Polarity'': When the elements of a row of Pascal's triangle are alternately added and subtracted together, the result is 0. For example, row 6 is 1,&nbsp;6,&nbsp;15,&nbsp;20,&nbsp;15,&nbsp;6,&nbsp;1, so the formula is 1&nbsp;−&nbsp;6&nbsp;+&nbsp;15&nbsp;−&nbsp;20&nbsp;+&nbsp;15&nbsp;−&nbsp;6&nbsp;+&nbsp;1&nbsp;=&nbsp;0.

=== Diagonals ===
[[File:Pascal_triangle_simplex_numbers.svg|thumb|upright=1.25|Derivation of [[simplex]] numbers from a left-justified Pascal's triangle]]
The diagonals of Pascal's triangle contain the [[Figurate numbers#Triangular numbers and their analogs in higher dimensions|figurate numbers]] of simplices:
* The diagonals going along the left and right edges contain only 1's.
* The diagonals next to the edge diagonals contain the [[natural number]]s in order. The 1-dimensional simplex numbers increment by 1 as the line segments extend to the next whole number along the number line.
* Moving inwards, the next pair of diagonals contain the [[triangular number]]s in order.
* The next pair of diagonals contain the [[tetrahedral number]]s in order, and the next pair give [[pentatope number]]s.

::<math>\begin{align}
P_0(n) &= P_d(0) = 1, \\
P_d(n) &= P_d(n-1) + P_{d-1}(n) \\
&= \sum_{i=0}^n P_{d-1}(i) = \sum_{i=0}^d P_i(n-1).
\end{align}</math>

The symmetry of the triangle implies that the ''n''<sup>th</sup> d-dimensional number is equal to the ''d''<sup>th</sup> ''n''-dimensional number.

An alternative formula that does not involve recursion is
<math display="block">P_d(n)=\frac{1}{d!}\prod_{k=0}^{d-1} (n+k) = {n^{(d)}\over d!} = \binom{n+d-1}{d},</math>
where ''n''<sup>(''d'')</sup> is the [[rising factorial]].

The geometric meaning of a function ''P''<sub>''d''</sub> is: ''P''<sub>''d''</sub>(1) = 1 for all ''d''. Construct a ''d''-[[dimensional]] triangle (a 3-dimensional [[triangle]] is a [[tetrahedron]]) by placing additional dots below an initial dot, corresponding to ''P''<sub>''d''</sub>(1) = 1. Place these dots in a manner analogous to the placement of numbers in Pascal's triangle. To find P<sub>''d''</sub>(''x''), have a total of ''x'' dots composing the target shape. P<sub>''d''</sub>(''x'') then equals the total number of dots in the shape. A 0-dimensional triangle is a point and a 1-dimensional triangle is simply a line, and therefore ''P''<sub>0</sub>(''x'') = 1 and ''P''<sub>1</sub>(''x'') = ''x'', which is the sequence of natural numbers. The number of dots in each layer corresponds to ''P''<sub>''d''&nbsp;−&nbsp;1</sub>(''x'').

=== Calculating a row or diagonal by itself ===
There are simple algorithms to compute all the elements in a row or diagonal without computing other elements or factorials.

To compute row <math>n</math> with the elements <math>\tbinom{n}{0}, \tbinom{n}{1}, \ldots, \tbinom{n}{n}</math>, begin with <math>\tbinom{n}{0}=1</math>. For each subsequent element, the value is determined by multiplying the previous value by a fraction with slowly changing numerator and denominator:

:<math> {n\choose k}= {n\choose k-1}\times \frac{n+1-k}{k}.</math>

For example, to calculate row 5, the fractions are&nbsp; <math>\tfrac{5}{1}</math>,&nbsp; <math>\tfrac{4}{2}</math>,&nbsp; <math>\tfrac{3}{3}</math>,&nbsp; <math>\tfrac{2}{4}</math> and <math>\tfrac{1}{5}</math>, and hence the elements are&nbsp; <math>\tbinom{5}{0}=1</math>, &nbsp; <math>\tbinom{5}{1}=1\times\tfrac{5}{1}=5</math>, &nbsp; <math>\tbinom{5}{2}=5\times\tfrac{4}{2}=10</math>, etc. (The remaining elements are most easily obtained by symmetry.)

To compute the diagonal containing the elements <math>\tbinom{n}{0}, \tbinom{n+1}{1}, \tbinom{n+2}{2},\ldots,</math> begin again with <math>\tbinom{n}{0} = 1</math> and obtain subsequent elements by multiplication by certain fractions:

:<math> {n+k\choose k}= {n+k-1\choose k-1}\times \frac{n+k}{k}.</math>

For example, to calculate the diagonal beginning at <math>\tbinom{5}{0}</math>, the fractions are&nbsp; <math>\tfrac{6}{1}, \tfrac{7}{2}, \tfrac{8}{3}, \ldots</math>, and the elements are <math>\tbinom{5}{0}=1, \tbinom{6}{1}=1 \times \tfrac{6}{1}=6, \tbinom{7}{2}=6\times\tfrac{7}{2}=21</math>, etc. By symmetry, these elements are equal to <math>\tbinom{5}{5}, \tbinom{6}{5}, \tbinom{7}{5}</math>, etc.

[[File:Fibonacci in Pascal triangle.png|thumb|[[Fibonacci sequence]] in Pascal's triangle]]

=== Overall patterns and properties ===
[[File:Sierpinski Pascal triangle.svg|thumb|A level-4 approximation to a [[Sierpiński triangle]] obtained by shading the first 32 rows of a Pascal triangle white if the binomial coefficient is even and black if it is odd.]]
* The pattern obtained by coloring only the odd numbers in Pascal's triangle closely resembles the [[fractal]] known as the [[Sierpiński triangle]]. This resemblance becomes increasingly accurate as more rows are considered; in the limit, as the number of rows approaches infinity, the resulting pattern ''is'' the Sierpiński triangle, assuming a fixed perimeter. More generally, numbers could be colored differently according to whether or not they are multiples of 3, 4, etc.; this results in other similar patterns.
:As the proportion of black numbers tends to zero with increasing ''n'', a corollary is that the proportion of odd binomial coefficients tends to zero as ''n'' tends to infinity.<ref>Ian Stewart, "How to Cut a Cake", Oxford University Press, page 180</ref>
<div class="thumb tright" style="clear: right; text-align: center;">
<div class="thumbinner" style="width: 230px;">
{| cellpadding="0" cellspacing="0" style="line-height: 0; background: white; font-size: 88%; border: 1px #b0b0b0 solid; padding: 0; margin: auto"
|- style="vertical-align: middle"
| style="padding: 0; vertical-align: inherit; background-color: #ffce9e;" | [[File:Chess rll45.svg|26x26px|alt=a4 white rook|a4 white rook]]
| style="padding: 0; vertical-align: inherit; background-color: #d18b47;" | [[File:Chess x1d45.svg|26x26px|alt=b4 one|b4 one]]
| style="padding: 0; vertical-align: inherit; background-color: #ffce9e;" | [[File:Chess x1l45.svg|26x26px|alt=c4 one|c4 one]]
| style="padding: 0; vertical-align: inherit; background-color: #d18b47;" | [[File:Chess x1d45.svg|26x26px|alt=d4 one|d4 one]]
|- style="vertical-align: middle"
| style="padding: 0; vertical-align: inherit; background-color: #d18b47;" | [[File:Chess x1d45.svg|26x26px|alt=a3 one|a3 one]]
| style="padding: 0; vertical-align: inherit; background-color: #ffce9e;" | [[File:Chess x2l45.svg|26x26px|alt=b3 two|b3 two]]
| style="padding: 0; vertical-align: inherit; background-color: #d18b47;" | [[File:Chess x3d45.svg|26x26px|alt=c3 three|c3 three]]
| style="padding: 0; vertical-align: inherit; background-color: #ffce9e;" | [[File:Chess x4l45.svg|26x26px|alt=d3 four|d3 four]]
|- style="vertical-align: middle"
| style="padding: 0; vertical-align: inherit; background-color: #ffce9e;" | [[File:Chess x1l45.svg|26x26px|alt=a2 one|a2 one]]
| style="padding: 0; vertical-align: inherit; background-color: #d18b47;" | [[File:Chess x3d45.svg|26x26px|alt=b2 three|b2 three]]
| style="padding: 0; vertical-align: inherit; background-color: #ffce9e;" | [[File:Chess x6l45.svg|26x26px|alt=c2 six|c2 six]]
| style="padding: 0; vertical-align: inherit; background-color: #d18b47;" | {{resize|150%|10}}
|- style="vertical-align: middle"
| style="padding: 0; vertical-align: inherit; background-color: #d18b47;" | [[File:Chess x1d45.svg|26x26px|alt=a1 one|a1 one]]
| style="padding: 0; vertical-align: inherit; background-color: #ffce9e;" | [[File:Chess x4l45.svg|26x26px|alt=b1 four|b1 four]]
| style="padding: 0; vertical-align: inherit; background-color: #d18b47;" | {{resize|150%|10}}
| style="padding: 0; vertical-align: inherit; background-color: #ffce9e;" | {{resize|150%|20}}
|}
<div class="thumbcaption">
Pascal's triangle overlaid on a grid gives the number of distinct paths to each square, assuming only rightward and downward steps to an adjacent square are considered.
</div></div></div>

* In a triangular portion of a grid (as in the images below), the number of shortest grid paths from a given node to the top node of the triangle is the corresponding entry in Pascal's triangle. On a [[Plinko]] game board shaped like a triangle, this distribution should give the probabilities of winning the various prizes.

[[Image:Pascal's Triangle 4 paths.svg|center|400px]]

* If the rows of Pascal's triangle are left-justified, the diagonal bands (colour-coded below) sum to the [[Fibonacci number]]s.
::{| style="align:center;"
|- align=center
|bgcolor=red|1
|- align=center
| style="background:orange;"|1
| style="background:yellow;"|1
|- align=center
| style="background:yellow;"|1
|bgcolor=lime|2
|bgcolor=aqua|1
|- align=center
|bgcolor=lime|1
|bgcolor=aqua|3
| style="background:violet;"|3
|bgcolor=red|1
|- align=center
|bgcolor=aqua|1
| style="background:violet;"|4
|bgcolor=red|6
| style="background:orange;"|4
| style="background:yellow;"|1
|- align=center
| style="background:violet;"|1
|bgcolor=red|5
| style="background:orange;"|10
| style="background:yellow;"|10
|bgcolor=lime|5
|bgcolor=aqua|1
|- align=center
|bgcolor=red|1
| style="background:orange;"|6
| style="background:yellow;"|15
|bgcolor=lime|20
|bgcolor=aqua|15
| style="background:violet;"|6
|bgcolor=red|1
|- align=center
| style="background:orange; width:40px;"|1
| style="background:yellow; width:40px;"|7
| style="background:lime; width:40px;"|21
| style="background:aqua; width:40px;"|35
| style="background:violet; width:40px;"|35
| style="background:red; width:40px;"|21
| style="background:orange; width:40px;"|7
| style="background:yellow; width:40px;"|1
|}

=== Construction as matrix exponential ===
{{Image frame|caption=Binomial matrix as matrix exponential. All the dots represent 0.
|content=<math>\begin{align}
\exp\begin{pmatrix}
. & . & . & . & . \\
1 & . & . & . & . \\
. & 2 & . & . & . \\
. & . & 3 & . & . \\
. & . & . & 4 & .
\end{pmatrix} &=
\begin{pmatrix}
1 & . & . & . & . \\
1 & 1 & . & . & . \\
1 & 2 & 1 & . & . \\
1 & 3 & 3 & 1 & . \\
1 & 4 & 6 & 4 & 1
\end{pmatrix}\\
e^{\text{counting}} &= \text{binomial}
\end{align}
</math>
}}
{{See also|Pascal matrix}}
Due to its simple construction by factorials, a very basic representation of Pascal's triangle in terms of the [[matrix exponential]] can be given: Pascal's triangle is the exponential of the matrix which has the sequence 1, 2, 3, 4, ... on its sub-diagonal and zero everywhere else.

===Construction of Clifford algebra using simplices===
Labelling the elements of each n-simplex matches the basis elements of [[Clifford algebra]] used as forms in [[Geometric_algebra|Geometric Algebra]] rather than matrices. Recognising the geometric operations, such as rotations, allows the algebra operations to be discovered. Just as each row, {{mvar|n}}, starting at 0, of Pascal's triangle corresponds to an {{mvar|(n-1)}}-simplex, as described below, it also defines the number of named basis forms in {{mvar|n}} dimensional [[Geometric algebra]]. The [[binomial theorem]] can be used to prove the geometric relationship provided by Pascal's triangle.<ref>{{citation |url=https://github.com/GPWilmot/geoalg|title=The Algebra Of Geometry|year=2023|last=Wilmot|first=G.P.}}</ref> This same proof could be applied to simplices except that the first column of all 1's must be ignored whereas in the algebra these correspond to the real numbers, <math>\R</math>, with basis 1.

=== Relation to geometry of polytopes ===
{{more sources|section|date=February 2025}}
Each row of Pascal's triangle gives the number of elements (such as edges and corners) of each dimension in a corresponding [[simplex]] (such as a triangle or tetrahedron).  In particular, for {{math| ''k'' > 0}}, the {{mvar|k}}th entry in the {{mvar|n}}th row is the number of {{math|(''k'' − 1)}}-dimensional elements in a {{math|(''n'' − 1)}}-dimensional simplex.  For example, a triangle (the 2-dimensional simplex) one 2-dimensional element (itself), three 1-dimensional elements (lines, or edges), and three 0-dimensional elements ([[Vertex (graph theory)|vertices]], or corners); this corresponds to the third row 1, 3, 3, 1 of Pascal's triangle.  This fact can be explained by combining Pascal's rule for generating the triangle with the geometric construction of simplices: each simplex is formed from a simplex of one lower dimension by the addition of a new vertex, outside the space in which the lower-dimensional simplex lies.  Then each {{mvar|d}}-dimensional element in the smaller simplex remains a {{mvar|d}}-dimensional element of the higher simplex, and each {{math|(''d'' − 1)}}-dimensional element when joined to the new vertex forms a new {{mvar|d}}-dimensional element of the higher simplex.<ref>{{Cite book |last=Coxeter |first=Harold Scott Macdonald |url=https://books.google.com/books?id=iWvXsVInpgMC |title=Regular Polytopes |date=1973-01-01 |publisher=Courier Corporation |isbn=978-0-486-61480-9 |edition=3rd |pages=118–144 |language=en |chapter=Chapter VII: ordinary polytopes in higher space, 7.2: Pyramids, dipyramids and prisms}}</ref>

A similar pattern is observed relating to [[square (geometry)|squares]], as opposed to triangles. To find the pattern, one must construct an analog to Pascal's triangle, whose entries are the coefficients of {{math|(''x'' + 2)<sup>row number</sup>}}, instead of {{math|(''x'' + 1)<sup>row number</sup>}}. There are a couple ways to do this. The simpler is to begin with row 0 = 1 and row 1 = 1, 2. Proceed to construct the analog triangles according to the following rule:

:<math> {n \choose k} = 2\times{n-1 \choose k-1} + {n-1 \choose k}.</math>

That is, choose a pair of numbers according to the rules of Pascal's triangle, but double the one on the left before adding. This results in:
:<math>\begin{matrix}
                           \text{  1} \\
                       \text{  1}   \quad \text{  2} \\
                   \text{  1}   \quad \text{  4}   \quad \text{  4} \\
               \text{  1}   \quad\text{  6}   \quad   \text{ 12}   \quad\text{  8} \\
           \text{  1}   \quad\text{  8}   \quad   \text{ 24}   \quad  \text{ 32}   \quad  \text{ 16} \\
       \text{  1}   \quad   \text{ 10}   \quad  \text{ 40}   \quad  \text{ 80}   \quad  \text{ 80}    \quad  \text{ 32} \\
   \text{  1}   \quad  \text{ 12}   \quad  \text{ 60}   \quad  160   \quad 240  \quad  192   \quad   \text{ 64} \\
\text{  1}   \quad  \text{ 14}   \quad  \text{ 84}   \quad  280  \quad  560  \quad  672   \quad  448   \quad   128
\end{matrix}</math>
The other way of producing this triangle is to start with Pascal's triangle and multiply each entry by 2<sup>k</sup>, where k is the position in the row of the given number. For example, the 2nd value in row 4 of Pascal's triangle is 6 (the slope of 1s corresponds to the zeroth entry in each row). To get the value that resides in the corresponding position in the analog triangle, multiply 6 by {{math|1=2<sup>position number</sup> = 6 × 2<sup>2</sup> = 6 × 4 = 24}}. Now that the analog triangle has been constructed, the number of elements of any dimension that compose an arbitrarily dimensioned [[cube (geometry)|cube]] (called a [[hypercube]]) can be read from the table in a way analogous to Pascal's triangle. For example, the number of 2-dimensional elements in a 2-dimensional cube (a square) is one, the number of 1-dimensional elements (sides, or lines) is 4, and the number of 0-dimensional elements (points, or vertices) is 4. This matches the 2nd row of the table (1, 4, 4). A cube has 1 cube, 6 faces, 12 edges, and 8 vertices, which corresponds to the next line of the analog triangle (1, 6, 12, 8). This pattern continues indefinitely.

To understand why this pattern exists, first recognize that the construction of an ''n''-cube from an {{math|1=(''n'' − 1)}}-cube is done by simply duplicating the original figure and displacing it some distance (for a regular ''n''-cube, the edge length) [[orthogonal]] to the space of the original figure, then connecting each vertex of the new figure to its corresponding vertex of the original. This initial duplication process is the reason why, to enumerate the dimensional elements of an ''n''-cube, one must double the first of a pair of numbers in a row of this analog of Pascal's triangle before summing to yield the number below. The initial doubling thus yields the number of "original" elements to be found in the next higher ''n''-cube and, as before, new elements are built upon those of one fewer dimension (edges upon vertices, faces upon edges, etc.). Again, the last number of a row represents the number of new vertices to be added to generate the next higher ''n''-cube.

In this triangle, the sum of the elements of row ''m'' is equal to 3<sup>''m''</sup>. Again, to use the elements of row 4 as an example: {{math|1=1 + 8 + 24 + 32 + 16 = 81}}, which is equal to <math>3^4 = 81</math>.

==== Counting vertices in a cube by distance ====
Each row of Pascal's triangle gives the number of vertices at each distance from a fixed vertex in an ''n''-dimensional cube. For example, in three dimensions, the third row (1 3 3 1) corresponds to the usual three-dimensional [[cube]]: fixing a vertex ''V'', there is one vertex at distance 0 from ''V'' (that is, ''V'' itself), three vertices at distance 1, three vertices at distance {{sqrt|2}} and one vertex at distance {{sqrt|3}} (the vertex opposite ''V''). The second row corresponds to a square, while larger-numbered rows correspond to [[hypercube]]s in each dimension.

=== Fourier transform of sin(''x'')<sup>''n''+1</sup>/''x'' ===

As stated previously, the coefficients of (''x''&nbsp;+&nbsp;1)<sup>''n''</sup> are the nth row of the triangle. Now the coefficients of (''x''&nbsp;−&nbsp;1)<sup>''n''</sup> are the same, except that the sign alternates from +1 to −1 and back again. After suitable normalization, the same pattern of numbers occurs in the [[Fourier transform]] of sin(''x'')<sup>''n''+1</sup>/''x''. More precisely: if ''n'' is even, take the [[real part]] of the transform, and if ''n'' is odd, take the [[imaginary part]]. Then the result is a [[step function]], whose values (suitably normalized) are given by the ''n''th row of the triangle with alternating signs.<ref>For a similar example, see e.g. {{citation|title=Solvent suppression in Fourier transform nuclear magnetic resonance|first=P. J.|last=Hore|journal=Journal of Magnetic Resonance|year=1983|volume=55|issue=2|pages=283–300|doi=10.1016/0022-2364(83)90240-8|bibcode=1983JMagR..55..283H}}.</ref> For example, the values of the step function that results from:

:<math>\mathfrak{Re}\left(\text{Fourier} \left[  \frac{\sin(x)^5}{x}   \right]\right)</math>

compose the 4th row of the triangle, with alternating signs. This is a generalization of the following basic result (often used in [[electrical engineering]]):

:<math>\mathfrak{Re}\left(\text{Fourier} \left[ \frac{\sin(x)^1}{x}\right] \right)</math>

is the [[boxcar function]].<ref>{{citation|title=An Introduction to Digital Signal Processing|first=John H.|last=Karl|publisher=Elsevier|year=2012|isbn=9780323139595|page=110|url=https://books.google.com/books?id=9Dv1PClLZWIC&pg=PA110}}.</ref>  The corresponding row of the triangle is row 0, which consists of just the number&nbsp;1.

If n is [[congruence relation|congruent]] to 2 or to 3 mod 4, then the signs start with&nbsp;−1. In fact, the sequence of the (normalized) first terms corresponds to the powers of [[imaginary unit|i]], which cycle around the intersection of the axes with the unit circle in the complex plane: <math display="block">  +i,-1,-i,+1,+i,\ldots  </math>

== Extensions ==

Pascal's triangle may be extended upwards, above the 1 at the apex, preserving the additive property, but there is more than one way to do so.<ref>{{cite book
 | display-authors = etal
 | last = Hilton | first = P.
 | series = In International Series in Modern Applied Mathematics and Computer Science
 | pages = 89–102
 | title = Symmetry 2
 | chapter = Extending the binomial coefficients to preserve symmetry and pattern
 | publisher = Pergamon
 | year = 1989
 | doi = 10.1016/B978-0-08-037237-2.50013-1
 | isbn = 9780080372372 | chapter-url = https://www.sciencedirect.com/science/article/pii/B9780080372372500131
 }}.</ref>

=== To higher dimensions ===
Pascal's triangle has higher [[dimension]]al generalizations. The three-dimensional version is known as ''[[Pascal's pyramid]]'' or ''Pascal's tetrahedron'', while the general versions are known as ''[[Pascal's simplex|Pascal's simplices]]''.

=== To complex numbers ===

When the factorial function is defined as <math>z! = \Gamma(z + 1)</math>, Pascal's triangle can be extended beyond the integers to <math>\Complex</math>, since <math>\Gamma(z + 1)</math> is [[Meromorphic_function|meromorphic]] to the entire [[complex plane]].<ref>{{cite book
 | display-authors = etal
 | last = Hilton | first = P.
 | series = In International Series in Modern Applied Mathematics and Computer Science
 | pages = 100–102
 | title = Symmetry 2
 | chapter = Extending the binomial coefficients to preserve symmetry and pattern
 | publisher = Pergamon
 | year = 1989
 | doi = 10.1016/B978-0-08-037237-2.50013-1
 | isbn = 9780080372372 | chapter-url = https://www.sciencedirect.com/science/article/pii/B9780080372372500131
 }}.</ref>

=== To arbitrary bases ===
[[Isaac Newton]] once observed that the first five rows of Pascal's triangle, when read as the digits of an integer, are the corresponding powers of eleven. He claimed without proof that subsequent rows also generate powers of eleven.<ref>{{citation
 | last = Newton | first = Isaac
 | journal = The Mathematical Works of Isaac Newton
 | title = A Treatise of the Method of Fluxions and Infinite Series
 | pages = 1:31–33
 | year = 1736
 | quote = But these in the alternate areas, which are given, I observed were the same with the figures of which the several ascending powers of the number 11 consist, viz. <math>11^{0}</math>, <math>11^{1}</math>, <math>11^{2}</math>, <math>11^{3}</math>, <math>11^{4}</math>, etc. that is, first 1; the second 1, 1; the third 1, 2, 1; the fourth 1, 3, 3, 1; the fifth 1, 4, 6, 4, 1, and so on
 | url = https://www.loc.gov/item/42048007/
 }}.</ref> In 1964, Robert L. Morton presented the more generalized argument that each [[#Rows|row]] <math>n</math> can be read as a radix <math>a</math> numeral, where [[E (mathematical constant)|<math>\lim_{n \to \infty} 11^{n}_{a}</math>]] is the hypothetical terminal row, or [[Infinite_product|limit]], of the triangle, and the rows are its partial products.<ref>{{citation
 | last = Morton | first = Robert L.
 | issue = 6
 | journal = The Mathematics Teacher
 | pages = 392–394
 | title = Pascal's Triangle and powers of 11
 | volume = 57
 | year = 1964
 | doi = 10.5951/MT.57.6.0392
 | jstor = 27957091
 }}.</ref> He proved the entries of row <math>n</math>, when interpreted directly as a place-value numeral, correspond to the binomial expansion of <math>(a + 1)^n = 11^{n}_{a}</math>. More rigorous proofs have since been developed.<ref>{{citation
 | display-authors = etal
 | last = Arnold | first = Robert
 | journal = Proceedings of Undergraduate Mathematics Day
 | title = Newton's Unfinished Business: Uncovering the Hidden Powers of Eleven in Pascal's Triangle
 | year = 2004
 | url = https://ecommons.udayton.edu/mth_epumd/6/
 }}.</ref><ref>{{citation
 | display-authors = etal
 | last = Islam | first = Robiul
 | title = Finding any row of Pascal's triangle extending the concept of power of 11
 | year = 2020
 | url = https://www.researchgate.net/publication/341785706
 }}.</ref> To better understand the principle behind this interpretation, here are some things to recall about binomials:
* A radix <math>a</math> numeral in [[positional notation]] (e.g. <math>14641_{a}</math>) is a univariate polynomial in the variable <math>a</math>, where the [[Monomial#Degree|degree]] of the variable of the <math>i</math>th [[Addition#summand|term]] (starting with <math>i = 0</math>) is <math>i</math>. For example, <math>14641_{a} = 1 \cdot a^{4} + 4 \cdot a^{3} + 6 \cdot a^{2} + 4 \cdot a^{1} + 1 \cdot a^{0}</math>.
* A row corresponds to the binomial expansion of <math>(a + b)^{n}</math>. The variable <math>b</math> can be eliminated from the expansion by setting <math>b = 1</math>. The expansion now typifies the expanded form of a radix <math>a</math> numeral,<ref>{{citation
 | last = Winteridge | first = David J.
 | issue = 1
 | journal = Mathematics in School
 | pages = 12–13
 | title = Pascal's Triangle and Powers of 11
 | volume = 13
 | year = 1984
 | jstor = 30213884
 }}.</ref><ref>{{citation
 | last = Kallós | first = Gábor
 | issue = 1
 | journal = Annales Mathématiques
 | pages = 1–15
 | title = A generalization of Pascal's triangle using powers of base numbers
 | volume = 13
 | year = 2006
 | 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 = \{c - 1, -(c + 1)\} \;\mathrm{mod}\; 2c</math> with odd values of <math>n</math> [[Negative_number#Multiplication|yielding]] negative row products.<ref>{{cite book
 | display-authors = etal
 | last = Hilton | first = P.
 | series = In International Series in Modern Applied Mathematics and Computer Science
 | pages = 89–91
 | title = Symmetry 2
 | chapter = Extending the binomial coefficients to preserve symmetry and pattern
 | publisher = Pergamon
 | year = 1989
 | doi = 10.1016/B978-0-08-037237-2.50013-1
 | isbn = 9780080372372 | chapter-url = https://www.sciencedirect.com/science/article/pii/B9780080372372500131
 }}.</ref><ref>{{citation
 | last = Mueller | first = Francis J.
 | issue = 5
 | journal = The Mathematics Teacher
 | pages = 425–428
 | title = More on Pascal's Triangle and powers of 11
 | volume = 58
 | year = 1965
 | doi = 10.5951/MT.58.5.0425
 | jstor = 27957164
 }}.</ref><ref>{{citation
 | last = Low | first = Leone
 | issue = 5
 | journal = The Mathematics Teacher
 | pages = 461–463
 | title = Even more on Pascal's Triangle and Powers of 11
 | volume = 59
 | year = 1966
 | doi = 10.5951/MT.59.5.0461
 | jstor = 27957385
 }}.</ref>

By setting the row's radix (the variable <math>a</math>) equal to one and ten, row <math>n</math> becomes the product <math>11^{n}_{1} = 2^{n}</math> and <math>11^{n}_{10} = 11^{n}</math>, respectively. To illustrate, consider <math>a = n</math>, which yields the row product <math>\textstyle n^n \left( 1 + \frac{1}{n} \right)^{n} = 11^{n}_{n}</math>. The numeric representation of [[Superparticular ratio|<math>11^{n}_{n}</math>]] is formed by concatenating the entries of row <math>n</math>. The twelfth row denotes the product:

: <math>11^{12}_{12} = 1:10:56:164:353:560:650:560:353:164:56:10:1_{12} = 27433a9699701_{12}</math>

with compound digits (delimited by ":") in radix twelve. The digits from <math>k = n - 1</math> through <math>k = 1</math> are compound because these row entries compute to values greater than or equal to twelve. To [[Trapdoor_function|normalize]]<ref>{{citation
 | last = Fjelstad | first = P.
 | issue = 9
 | journal = Computers & Mathematics with Applications
 | pages = 3
 | title = Extending Pascal's Triangle 
 | volume = 21
 | doi = 10.1016/0898-1221(91)90119-O
 | year = 1991
 | doi-access = free
 }}.</ref> the numeral, simply carry the first compound entry's prefix, that is, remove the prefix of the coefficient <math>\textstyle {n \choose n - 1}</math> from its leftmost digit up to, but excluding, its rightmost digit, and use radix-twelve arithmetic to sum the removed prefix with the entry on its immediate left, then repeat this process, proceeding leftward, until the leftmost entry is reached. In this particular example, the normalized string ends with <math>01</math> for all <math>n</math>. The leftmost digit is <math>2</math> for <math>n > 2</math>, which is obtained by carrying the <math>1</math> of <math>10_{n}</math> at entry <math>k = 1</math>. It follows that the length of the normalized value of <math>11^{n}_{n}</math> is [[Bijection|equal]] to the row length, <math>n + 1</math>. The integral part of <math>1.1^{n}_{n}</math> contains exactly one digit because <math>n</math> (the number of places to the left the decimal has moved) is one less than the row length. Below is the normalized value of <math>1.1^{1234}_{1234}</math>. Compound digits remain in the value because they are radix <math>1234</math> [[Modular_arithmetic#Integers modulo m|residues]] represented in radix ten: 

: <math>1.1^{1234}_{1234} = 2.885:2:35:977:696:\overbrace{\ldots}^\text{1227 digits}:0:1_{1234} = 2.717181235\ldots_{10}</math>

== See also ==

{{div col|colwidth=20em}}
* [[Bean machine]], Francis Galton's "quincunx"
* [[Bell triangle]]
* [[Bernoulli's triangle]]
* [[Binomial expansion]]
* [[Cellular automata]]
* [[Euler triangle]]
* [[Floyd's triangle]]
* [[Gaussian binomial coefficient]]
* [[Hockey-stick identity]]
* [[Leibniz harmonic triangle]]
* [[Multiplicities of entries in Pascal's triangle]] (Singmaster's conjecture)
* [[Pascal matrix]]
* [[Pascal's pyramid]]
* [[Pascal's simplex]]
* [[Proton NMR]], one application of Pascal's triangle
* [[Star of David theorem]]
* [[Trinomial expansion]]
* [[Trinomial triangle]]
* [[Polynomials calculating sums of powers of arithmetic progressions]]
{{div col end}}

==References==
{{Reflist}}

== External links ==
* {{springer|title=Pascal triangle|id=p/p071790}}
* {{MathWorld | urlname=PascalsTriangle | title=Pascal's triangle }}
* [https://york.ac.uk/depts/maths/histstat/images/triangle.gif The Old Method Chart of the Seven Multiplying Squares] ''(from the Ssu Yuan Yü Chien of Chu Shi-Chieh, 1303, depicting the first nine rows of Pascal's triangle)''
* [https://web.archive.org/web/20040803130916/https://lib.cam.ac.uk/RareBooks/PascalTraite/ Pascal's Treatise on the Arithmetic Triangle] ''(page images of Pascal's treatise, 1654; [https://web.archive.org/web/20040803233048/https://lib.cam.ac.uk/RareBooks/PascalTraite/pascalintro.pdf summary])''

{{Blaise Pascal}}
{{Authority control}}

{{DEFAULTSORT:Pascal's triangle}}
[[Category:Factorial and binomial topics]]
[[Category:Blaise Pascal]]
[[Category:Triangles of numbers]]