Difference between revisions of "Pascal's triangle"

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== Definition ==
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'''Pascal's triangle''' is a triangle which contains the values from the [[binomial expansion]]; its various properties play a large role in [[combinatorics]].
  
Pascal's triangle is a triangle in which every number is the sum of the two numbers directly above it. The first few rows of Pascal's Triangle are depicted above. 
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== Properties ==
[[Image:Pascal.hex2.gif|left|thumb|400px|''[[Pascal's Triangle]]'' ]]
 
  
== Uses ==
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=== Binomial coefficients ===
Pascal's triangle has many interesting uses and applications.
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[[Image:First 9 rows of pascal's.png|frame|right|200px|These are the first nine rows of Pascal's Triangle.]] Pascal's Triangle is defined such that the number in row <math>n</math> and column <math>k</math> is <math>{n\choose k}</math>.  For this reason, [[mathematical convention|convention]] holds that both row numbers and column numbers start with 0.  Thus, the apex of the triangle is row 0, and the first number in each row is column 0.  As an example, the number in row 4, column 2 is <math>{4 \choose 2} = 6</math>.  Pascal's Triangle thus can serve as a "look-up table" for binomial expansion values.  Also, many of the characteristics of Pascal's Triangle are derived from [[combinatorial identities]]; for example, because <math>\sum_{k=0}^{n}{{n \choose k}}=2^n</math>, the sum of the values on row <math>n</math> of Pascal's Triangle is <math>2^n</math>.
  
=== Binomial Coefficients ===
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=== Sum of previous values ===
It can be noted that the numbers in Pascal's triangles are the [[binomial coefficients]].  This means that every number in Pascal's Triangle can be expressed as <math>\displaystyle{n\choose k-1}</math>, where n is the row number, and k is the place in the row.  For example, <math>\displaystyle{5\choose 3}</math> would represent the 5th row, and the 4th number in it.  It should be noted here that the first row of Pascal's Triangle is considered to be the row 1, 1, meaning that the "[[apex]]" of Pascal's Triangle is considered "Row 0."
 
  
=== Counting & Probability ===
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One of the best known features of Pascal's Triangle is derived from the combinatorics identity <math>{n \choose k}+{n \choose k+1} = {n+1 \choose k+1}</math>.  Thus, any number in the interior of Pascal's Triangle will be the sum of the two numbers appearing above it.  For example, <math>{5 \choose 1}+{5 \choose 2} = 5 + 10 = 15 = {6 \choose 2}</math>.  This property allows the easy creation of the first few rows of Pascal's Triangle without having to calculate out each binomial expansion.
If we flip x coins, and want to know what the probability is that y of them are heads, then we go to the xth row, and find the (y+1)th number (call this a).  Then, we find the sum of the numbers in the xth row (call this b).  Then, the probability is simply <math>\frac{a}{b}</math>
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=== Fibonacci numbers ===
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The [[Fibonacci sequence|Fibonacci numbers]] appear in Pascal's Triangle along the "shallow diagonals."  That is, <math>{n \choose 0}+{n-1 \choose 1}+\cdots+{n-\left\lfloor\frac{n}{2}\right\rfloor \choose \left\lfloor \frac{n}{2}\right \rfloor} = F(n+1)</math>, where <math>F(n)</math> is the Fibonacci sequence.  For example, <math>{6 \choose 0}+{5 \choose 1}+{4 \choose 2}+{3 \choose 3} = 1 + 5 + 6 + 1 = 13 = F(7)</math>.  A "shallow diagonal" is plotted in the diagram.
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=== Hockey-stick theorem ===
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The [[Combinatorial identity#Hockey-Stick Identity|Hockey-stick theorem]] states:
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<math>{n \choose 0}+{n+1 \choose 1}+\cdots+{n+k \choose k} = {n+k+1 \choose k}</math>.  Its name is due to the "hockey-stick" which appears when the numbers are plotted on Pascal's Triangle, as shown in the representation of the theorem below (where <math>n=2</math> and <math>k=3</math>).
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<asy>
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int chew(int n,int r){
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int res=1;
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for(int i=0;i<r;++i){
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  res=quotient(res*(n-i),i+1);
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  }
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return res;
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}
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for(int n=0;n<9;++n){
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for(int i=0;i<=n;++i){
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  if((i==2 && n<6)||(i==3 && n==6)){
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  if(n==6){label(string(chew(n,i)),(11+n/2-i,-n),p=red+2.5);}
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  else{label(string(chew(n,i)),(11+n/2-i,-n),p=blue+2);}
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  }
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  else{
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  label(string(chew(n,i)),(11+n/2-i,-n));
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  }
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  }
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}
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</asy>
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=== Number Parity ===
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Consider writing the row number <math>n</math> in [[base number | base]] two as <math>({n})_{10} = {(a_xa_{x-1} \cdots a_1a_0)}_2</math><math> = a_x 2^x+a_{x-1} 2^{x-1}+\cdots+a_1 2^1+a_0 2^0</math>The number in the <math>k</math>th column of the <math>n</math>th row in Pascal's Triangle is [[odd integer | odd]] if and only if <math>k</math> can be expressed as the sum of some <math>a_i 2^i</math>.  For example, <math>(9)_{10} = {(1001)}_{2} = 2^{3}+2^{0}</math>.  Thus, the only 4 odd numbers in the 9th row will be in the <math>{(0000)}_{2} = 0</math>th, <math>{(0001)}_{2} = 2^0 = 1</math>st, <math>{(1000)}_{2} = 2^3 = 8</math>th, and <math>{(1001)}_{2} = 2^3+2^0 = 9</math>th columnsAdditionally, marking each of these odd numbers in Pascal's Triangle creates a [[Sierpinski triangle]].
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[[Image:Sierpinski.jpg]]
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=== Patterns and Properties of the Pascal's Triangle ===
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==== Rows ====
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The zeroth row has a sum of <math> 1=2^0 </math>. The first row has a sum of <math> 2=2^1 </math>. The <math>n^{th} </math> row has a sum of <math> 2^n </math>
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=== Diagonals ===
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The 1st downward diagonal is a row of 1's, the 2nd downward diagonal on each side consists of the [[natural numbers]], the 3rd diagonal the [[triangular numbers]], and the 4th the [[pyramidal numbers]].
  
 
==See Also==
 
==See Also==
 
*[[Binomial theorem]]
 
*[[Binomial theorem]]
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*[[Pascal Triangle Related Problems]]
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[[Category:Combinatorics]]
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[[Category:Definition]]

Latest revision as of 12:47, 11 March 2019

Pascal's triangle is a triangle which contains the values from the binomial expansion; its various properties play a large role in combinatorics.

Properties

Binomial coefficients

These are the first nine rows of Pascal's Triangle.

Pascal's Triangle is defined such that the number in row $n$ and column $k$ is ${n\choose k}$. For this reason, convention holds that both row numbers and column numbers start with 0. Thus, the apex of the triangle is row 0, and the first number in each row is column 0. As an example, the number in row 4, column 2 is ${4 \choose 2} = 6$. Pascal's Triangle thus can serve as a "look-up table" for binomial expansion values. Also, many of the characteristics of Pascal's Triangle are derived from combinatorial identities; for example, because $\sum_{k=0}^{n}{{n \choose k}}=2^n$, the sum of the values on row $n$ of Pascal's Triangle is $2^n$.

Sum of previous values

One of the best known features of Pascal's Triangle is derived from the combinatorics identity ${n \choose k}+{n \choose k+1} = {n+1 \choose k+1}$. Thus, any number in the interior of Pascal's Triangle will be the sum of the two numbers appearing above it. For example, ${5 \choose 1}+{5 \choose 2} = 5 + 10 = 15 = {6 \choose 2}$. This property allows the easy creation of the first few rows of Pascal's Triangle without having to calculate out each binomial expansion.

Fibonacci numbers

The Fibonacci numbers appear in Pascal's Triangle along the "shallow diagonals." That is, ${n \choose 0}+{n-1 \choose 1}+\cdots+{n-\left\lfloor\frac{n}{2}\right\rfloor \choose \left\lfloor \frac{n}{2}\right \rfloor} = F(n+1)$, where $F(n)$ is the Fibonacci sequence. For example, ${6 \choose 0}+{5 \choose 1}+{4 \choose 2}+{3 \choose 3} = 1 + 5 + 6 + 1 = 13 = F(7)$. A "shallow diagonal" is plotted in the diagram.

Hockey-stick theorem

The Hockey-stick theorem states: ${n \choose 0}+{n+1 \choose 1}+\cdots+{n+k \choose k} = {n+k+1 \choose k}$. Its name is due to the "hockey-stick" which appears when the numbers are plotted on Pascal's Triangle, as shown in the representation of the theorem below (where $n=2$ and $k=3$).

[asy] int chew(int n,int r){  int res=1;  for(int i=0;i<r;++i){   res=quotient(res*(n-i),i+1);   }  return res;  } for(int n=0;n<9;++n){  for(int i=0;i<=n;++i){   if((i==2 && n<6)||(i==3 && n==6)){    if(n==6){label(string(chew(n,i)),(11+n/2-i,-n),p=red+2.5);}    else{label(string(chew(n,i)),(11+n/2-i,-n),p=blue+2);}    }   else{    label(string(chew(n,i)),(11+n/2-i,-n));    }   }  } [/asy]

Number Parity

Consider writing the row number $n$ in base two as $({n})_{10} = {(a_xa_{x-1} \cdots a_1a_0)}_2$$= a_x 2^x+a_{x-1} 2^{x-1}+\cdots+a_1 2^1+a_0 2^0$. The number in the $k$th column of the $n$th row in Pascal's Triangle is odd if and only if $k$ can be expressed as the sum of some $a_i 2^i$. For example, $(9)_{10} = {(1001)}_{2} = 2^{3}+2^{0}$. Thus, the only 4 odd numbers in the 9th row will be in the ${(0000)}_{2} = 0$th, ${(0001)}_{2} = 2^0 = 1$st, ${(1000)}_{2} = 2^3 = 8$th, and ${(1001)}_{2} = 2^3+2^0 = 9$th columns. Additionally, marking each of these odd numbers in Pascal's Triangle creates a Sierpinski triangle.

Sierpinski.jpg

Patterns and Properties of the Pascal's Triangle

Rows

The zeroth row has a sum of $1=2^0$. The first row has a sum of $2=2^1$. The $n^{th}$ row has a sum of $2^n$

Diagonals

The 1st downward diagonal is a row of 1's, the 2nd downward diagonal on each side consists of the natural numbers, the 3rd diagonal the triangular numbers, and the 4th the pyramidal numbers.

See Also

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