Explanation: Pascal's Triangle (Triangle of Binomial Coefficients)

(related to Definition: Combinations)

The numbers formed by binomial coefficients \(\binom nk\), where \(n\) and \(k\) are natural numbers form a scheme, named the Pascal's triangle1. The beginning of Pascal's triangle (for \(11\le n,k\le 0\)) is shown in the following table:

\[\begin{array}{r|rrrrrrrrrrr} n&\binom n0&\binom n1&\binom n2&\binom n3&\binom n4&\binom n5&\binom n6&\binom n7&\binom n8&\binom n9&\binom n{10}&\binom n{11}\\ \hline 0&1\\ 1&1&1&\\ 2&1&2&1&\\ 3&1&3&3&1&\\ 4&1&4&6&4&1\\ 5&1&5&10&10&5&1\\ 6&1&6&15&20&15&6&1\\ 7&1&7&21&35&35&21&7&1\\ 8&1&8&28&56&70&56&28&8&1\\ 9&1&9&36&84&126&126&84&36&9&1\\ 10&1&10&45&120&210&252&210&120&45&10&1\\ 11&1&11&55&165&330&462&462&330&165&55&11&1\\ \end{array}\]

Note that empty entries in this table are actually \(0\)'s. This is because the numerator in the closed formula for binomial coefficients is zero if \(n < k\) \[\binom nk=\frac{n(n-1)\cdots(n-k+1)}{k(k-1)\cdots 2\cdot 1},\] e.g. \(\binom 23=(1\cdot 0\cdot (-1))/(3\cdot 2\cdot 1)=0\). The \(0\) entries have been left blank to help emphasize the rest of the table.

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Footnotes


  1. named after Blaise Pascal (1623-1662)