= Pascal's triangle
{c}
{title2=$T_{n,k}=\binom nk$}
{wiki}
= Pascal triangle
{c}
{synonym}
Row $n$ of Pascal's triangle consists of the <binomial coefficients> $\binom n0,\ldots,\binom nn$, beginning with row $0$. Boundary entries are one, and <Pascal's identity> says each interior entry is the sum of the two entries above it. The <hockey-stick identity> evaluates sums along its diagonals.
The highlighted entry in the <binomial coefficient> illustration is $\binom52=10$. Split two-person teams according to whether they contain the fifth person: six choose both members from the other four, and four contain the fifth person together with one of those four. Thus $10=6+4$. All entries can be generated by the same local rule, so a large table of counts needs no repeated enumeration of its <subsets>.
\b[Reading row seven.] Label the entries in each row from $k=0$ on the left through $k=n$ on the right. Rows zero through seven are shown below. The highlighted entry $T_{7,3}=35$ comes from $T_{6,2}=15$ and $T_{6,3}=20$. Starting with row zero $(1)$, the edges remain one and <Pascal's rule> determines every interior entry, so <mathematical induction> uniquely determines the whole array.
\Image[/pascal-triangle-seven-step-count.png]
{title=Rows zero through seven of <Pascal's triangle>, with $T_{7,3}=T_{6,2}+T_{6,3}=15+20=35$}
{description=The index $k$ increases from left to right in each row; it counts north steps in the <north-east lattice path> interpretation. Each row index $n$ counts all steps.}
The triangular and rectangular layouts describe the same entries. For a <north-east lattice path> ending at $(a,b)$, set $n=a+b$ and $k=b$, so $P(a,b)=T_{n,k}$. The predecessor $(a-1,b)$ has indices $(n-1,k)$, while $(a,b-1)$ has indices $(n-1,k-1)$. Thus the grid's west and south arrivals are exactly the triangle's two parents. This change of indices explains why the <north-east lattice path recurrence> and <Pascal's rule> agree.
There is also a global check on the local construction: row $n$ sums to $2^n$. Every <subset> of an $n$-element <set> appears exactly once among the classes of sizes $0,\ldots,n$, and each element has two inclusion choices by the <rule of product>. In row seven, $1+7+21+35+35+21+7+1=128=2^7$. Replacing every <subset> by its <complement> reverses the row, explaining $T_{n,k}=T_{n,n-k}$. The <binomial theorem> packages the whole row as the <coefficients> of $(1+x)^n$.
<Blaise Pascal>'s Traité du triangle arithmétique, published in 1665, presents an arithmetic array and connects it to combinations, games of chance, and powers of binomial expressions. This makes the connection between a counting table and several applications visible in the original work. https://archive.org/details/bub_gb_UqgUAAAAQAAJ[Pascal, Traité du triangle arithmétique, 1665 edition]{ref} The University of Cambridge's account of the treatise explains its 1654 composition and the organization of these applications. https://api.repository.cam.ac.uk/server/api/core/bitstreams/9bfbc9c8-fa56-4694-b96d-fc18db0f562f/content[Cambridge University Library, account of Pascal's treatise]{ref}
Back to article page
