132.76.50.5: /* Mass density */

2014-11-11T09:01:42Z

Mass density

← Older revision		Revision as of 11:01, 11 November 2014
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]~~]

en>Fgnievinski: /* See also */

2013-10-30T14:05:35Z

en>EmausBot: r2.7.2+) (Robot: Adding be:Канцэнтрацыя часціц

2012-07-13T20:37:04Z

r2.7.2+) (Robot: Adding be:Канцэнтрацыя часціц

Show changes

Number density - Revision history

132.76.50.5: /* Mass density */

en>Fgnievinski: /* See also */

en>EmausBot: r2.7.2+) (Robot: Adding be:Канцэнтрацыя часціц