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.

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, ${5 \choose 1}+{5 \choose 2} = 5 + 10 = 15 = {6 \choose 2}$ , as shown in the diagram. 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, ${n \choose 0}+{n-1 \choose 1}+\cdots+{n-\lfloor\frac{n}{2}\rfloor \choose \lfloor \frac{n}{2} \rfloor} = F(n+1)$ , 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.