|
|
| Line 1: |
Line 1: |
| {{distinguish|Pascal's law}}
| |
| In [[mathematics]], '''Pascal's rule''' is a [[combinatorics|combinatorial]] [[identity (mathematics)|identity]] about [[binomial coefficient]]s. It states that for any [[natural number]] ''n'' we have
| |
|
| |
|
| :<math>{n-1\choose k} + {n-1\choose k-1} = {n\choose k}\quad\text{for }1 \le k \le n </math>
| |
|
| |
|
| where <math>{n\choose k}</math> is a binomial coefficient. This is also commonly written
| | The author's title is Raul. His working day occupation is a production and preparing officer. North Dakota is exactly where my home is but his spouse wants them to transfer. Home [http://answers.yahoo.com/search/search_result?p=brewing&submit-go=Search+Y!+Answers brewing] is the thing I love most. He's been working on his web site for some time now. Verify it out here: http://dayzspain.com/wiki/index.php?title=Unanswered_Questions_on_Nya_Online_Casinon_P%C3%A5_N%C3%A4tet_That_You_Should_Know_About<br><br>my web site ... Nya online casinon 2014, [http://dayzspain.com/wiki/index.php?title=Unanswered_Questions_on_Nya_Online_Casinon_P%C3%A5_N%C3%A4tet_That_You_Should_Know_About Anbefales Web -site], |
| | |
| :<math>
| |
| {n \choose k} + {n \choose k-1} = {n + 1 \choose k} \quad\text{for } 1 \le k \le n + 1
| |
| </math>
| |
| | |
| ==Combinatorial proof==
| |
| [[Blaise Pascal|Pascal's]] rule has an intuitive combinatorial meaning. Recall that <math>{a\choose b}</math> counts in how many ways can we pick a [[subset]] with ''b'' elements out from a set with ''a'' elements. Therefore, the right side of the identity <math>{n\choose k}</math> is counting how many ways can we get a ''k''-subset out from a set with ''n'' elements.
| |
| | |
| Now, suppose you distinguish a particular element 'X' from the set with ''n'' elements. Thus, every time you choose ''k'' elements to form a subset there are two possibilities: ''X'' belongs to the chosen subset or not.
| |
| | |
| If ''X'' is in the subset, you only really need to choose ''k'' − 1 more objects (since it is known that ''X'' will be in the subset) out from the remaining ''n'' − 1 objects. This can be accomplished in <math>n-1\choose k-1</math> ways.
| |
| | |
| When ''X'' is not in the subset, you need to choose all the ''k'' elements in the subset from the ''n'' − 1 objects that are not ''X''. This can be done in <math>n-1\choose k</math> ways.
| |
| | |
| We conclude that the numbers of ways to get a ''k''-subset from the ''n''-set, which we know is <math>{n\choose k}</math>, is also the number
| |
| <math>{n-1\choose k-1} + {n-1\choose k}.</math>
| |
| | |
| See also [[Bijective proof]].
| |
| | |
| ==Algebraic proof==
| |
| We need to show
| |
| :<math> { n \choose k } + { n \choose k-1 } = { n+1 \choose k }.</math>
| |
| | |
| Let us begin by writing the left-hand side as
| |
| | |
| :<math> \frac{n!}{k!(n-k)!} + \frac{n!}{(k-1)!(n-(k-1))!}.</math>
| |
| | |
| Getting a common denominator and simplifying, we have
| |
| | |
| :<math>
| |
| \begin{align}
| |
| & {} \qquad \frac{n!}{k!(n-k)!} + \frac{n!}{(k-1)!(n-k+1)!} \\\\
| |
| & = \frac{(n-k+1)n!}{(n-k+1)k!(n-k)!}+\frac{kn!}{k(k-1)!(n-k+1)!} \\\\
| |
| & = \frac{(n-k+1)n!+kn!}{k!(n-k+1)!} \\\\ | |
| & = \frac{(n+1)n!}{k!((n+1)-k)!} \\\\
| |
| & = \frac{(n+1)!}{k!((n+1)-k)!} \\\\
| |
| & = { n+1 \choose k }.
| |
| \end{align}
| |
| </math>
| |
| | |
| ==Generalization==
| |
| Let <math>n, k_1, k_2, k_3,\dots ,k_p, p \in \mathbb{N}^* \,\!</math> and <math>n=k_1+k_2+k_3+ \cdots +k_p \,\!</math>. Then
| |
| | |
| : <math>
| |
| \begin{align}
| |
| & {} \quad {n-1\choose k_1-1,k_2,k_3, \dots, k_p}+{n-1\choose k_1,k_2-1,k_3,\dots, k_p}+\cdots+{n-1\choose k_1,k_2,k_3,\dots,k_p-1} \\
| |
| & = \frac{(n-1)!}{(k_1-1)!k_2!k_3! \cdots k_p!} + \frac{(n-1)!}{k_1!(k_2-1)!k_3!\cdots k_p!} + \cdots + \frac{(n-1)!}{k_1!k_2!k_3! \cdots (k_p-1)!} \\
| |
| & = \frac{k_1(n-1)!}{k_1!k_2!k_3! \cdots k_p!} + \frac{k_2(n-1)!}{k_1!k_2!k_3! \cdots k_p!} + \cdots + \frac{k_p(n-1)!}{k_1!k_2!k_3! \cdots k_p!} = \frac{(k_1+k_2+\cdots+k_p) (n-1)!}{k_1!k_2!k_3!\cdots k_p!} \\
| |
| & = \frac{n(n-1)!}{k_1!k_2!k_3! \cdots k_p!} = \frac{n!}{k_1!k_2!k_3! \cdots k_p!}
| |
| = {n\choose k_1, k_2, k_3, \dots , k_p}.
| |
| \end{align}
| |
| </math>
| |
| | |
| ==See also==
| |
| * [[Pascal's triangle]]
| |
| | |
| ==Sources==
| |
| *{{PlanetMath attribution|id=246|title=Pascal's rule}}
| |
| *{{PlanetMath attribution|id=259|title= Pascal's rule proof}}
| |
| *Merris, Russell. [http://media.wiley.com/product_data/excerpt/6X/04712629/047126296X.pdf''Combinatorics'']. John Wiley & Sons. 2003 ISBN 978-0-471-26296-1
| |
| | |
| ==External links==
| |
| * {{planetmath reference|id=5936|title=Central binomial coefficient}}
| |
| * {{planetmath reference|id=273|title=Binomial coefficient}}
| |
| * {{planetmath reference|id=4248|title=Pascal's triangle}}
| |
| | |
| {{DEFAULTSORT:Pascal's Rule}}
| |
| [[Category:Combinatorics]]
| |
| [[Category:Mathematical identities]]
| |
| [[Category:Articles containing proofs]]
| |
| | |
| [[ar:قاعدة باسكال]]
| |
| [[bg:Правило на Паскал]]
| |
| [[ca:Regla de Pascal]]
| |
| [[pt:Relação de Stifel]]
| |
| [[ru:Закон Паскаля]]
| |
| [[sv:Pascals identitet]]
| |
| [[zh:帕斯卡法則]]
| |
The author's title is Raul. His working day occupation is a production and preparing officer. North Dakota is exactly where my home is but his spouse wants them to transfer. Home brewing is the thing I love most. He's been working on his web site for some time now. Verify it out here: http://dayzspain.com/wiki/index.php?title=Unanswered_Questions_on_Nya_Online_Casinon_P%C3%A5_N%C3%A4tet_That_You_Should_Know_About
my web site ... Nya online casinon 2014, Anbefales Web -site,