Editing Pascal's triangle (section)

=== Relation to geometry of polytopes ===
{{more citations needed 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.