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Binomial Coefficients Revisited

The binomial coefficient \binom{n}{k} was originally defined in terms of the factorial notation, and with our recursive definitions of the factorial notation, we also have a complete and legally-correct definition of…

Binomial Coefficients Revisited

The binomial coefficient \(\binom{n}{k}\) was originally defined in terms of the factorial notation, and with our recursive definitions of the factorial notation, we also have a complete and legally-correct definition of binomial coefficients. The following recursive formula provides an efficient computational scheme.

Let \(n\) and \(k\) be integers with \(0\le k\le n\). If \(k=0\) or \(k=n\), set \(\binom{n}{k}=1\). If \(0\lt k\lt n\), set \[\begin{aligned}\end{aligned}\] This recursion has a natural combinatorial interpretation. Both sides count the number of \(k\)-element subsets of \(\{1,2,\dots,n\}\), with the right-hand side first grouping them into those which contain the element\(n\) and then those which don't. The traditional form of displaying this recursion is This pattern is called Pascal's triangle. Other than the \(1\)s at the ends of each row, an entry of the triangle is determined by adding the entry to the left and the entry to the right in the row above.

Xing was intrigued by the fact that he now had two fundamentally different ways to calculate binomial coefficients. One way is to write \(\binom{n}{m}=P(n,m)/m!\) and just carry out the specified arithmetic. The second way is to use the recursion of Pascal's triangle, so that you are just performing additions. So he experimented by writing a computer program to calculate binomial coefficients, using a library that treats big integers as strings. Which of the two ways do you think proved to be faster when \(n\) say was between \(1800\) and \(2000\) and \(m\) was around \(800\)?

Indi sen Hiç bir hasapçy buny çözmez, emma onuň bölekleri hasaplanmak üçindir. Aşakda birini synla ýa-da özüňi ýaz.

Tälim (1)

Her birini kagyzda synlap gör. Çözgüni barlap görmek üçin aç; barlap görlenler her bir ädim üçin çözgüççide açylyp bilner.

  1. Write a couple of sentences to explain why \(C(n-1,k-1)\) counts the number of \(k\)-element subsets of \(\{1,2,\dots,n\}\) that contain the element \(n\).

Özüň işle

Mugt hasap her bir ders üçin notlar goşýar, tamamlanan zadyňyzyň ýazgysyny, çözen problemaňyzy bir ýerde, we bu sahypa barada sorap biljek mugallymy. Matematika özüçe her kim üçin açyk, girseň ýa girmeseň.

_Gün _Giriş

Şu ýerde ulanylan simwollar

Eň täze düşündirme, surat, we her harpyň manysy üçin her bir simwoly basyň.

Adamlar soraýan soraglar

Permutation or combination?

Ask whether order matters. A lock code is a permutation (order matters); a hand of cards is a combination (it does not).

What is a graph in this sense?

Dots (vertices) joined by lines (edges), not a plot. Road maps, social networks and molecules are graphs; questions like "is there a route" and "how few colours" are graph theory.

Bu sahypanyň käbir bölegi şundan alyndy Keller & Trotter, Applied Combinatorics (CC BY-SA 4.0). Bu ýerde gysgaldylan we tekrar beýan edilen; hatalar biziňkidir.

_Ýaşa Combinatorics & Graph Theory