Pascal's triangle

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


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 Identity

The Hockey-Stick Identity 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.



This is a special case of Kummer's Theorem, which states that given a prime p and integers m,n, the highest power of p dividing $\binom{m}{n}$ is the number of carries in adding $m-n$ and n in base p.

Patterns and Properties of the Pascal's Triangle


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 $2^n$, which means there are $2^n$ ways to get to that row from the zeroth row.

Proof: To find the number of ways to get to the $n^{th}$ row of Pascal's Triangle, let's start with the zeroth row. There are $2$ ways to get to the next row because there are $2$ numbers in the first row. Now to get to the second row, we realize that starting from any number on the first row, we have two options to go to in the second row. This means we need to multiply the number of ways to get to the first row by $2$, which gives us $2 \cdot 2 = 2^2 = 4$. This pattern continues, so the number of ways to get to the $n^{th}$ row of Pascal's Triangle is $2^n$.


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