| Line 15: | Line 15: | ||
<math> = {n \choose n-r} </math>. | <math> = {n \choose n-r} </math>. | ||
QED | QED | ||
| + | |||
| + | ==Combinatorial identities== | ||
| + | <math> {n \choose k} = {n-1 \choose k} + {n-1 \choose k-1} </math> | ||
==Binomial Coefficients== | ==Binomial Coefficients== | ||
Revision as of 10:45, 7 September 2008
Contents
Permutations
Combinations
Claim
$ {n \choose r} = {n \choose n-r} $
Proof(Counting Argument)
$ {n \choose r} = $
$ = $ # ways to CHOOSE r things from n
$ = $ # ways of LEAVING 'n-r' things from n
$ = $ # ways of CHOOSING 'n-r' things from n $ = {n \choose n-r} $. QED
Combinatorial identities
$ {n \choose k} = {n-1 \choose k} + {n-1 \choose k-1} $
Binomial Coefficients
Pascal's Triangle
Here are the first five rows of Pascal's Triangle $ \begin{matrix} &&&&&1\\ &&&&1&&1\\ &&&1&&2&&1\\ &&1&&3&&3&&1\\ &1&&4&&6&&4&&1 \end{matrix} $
Binomial Theorem
Definition
$ (x+y)^n = \sum_{i=0}^n {n \choose k}x^i y^{n-i}, $
$ \text{where } {n \choose k} = \frac{n!}{n!(n-r)!}. $
Example
- What is $ \sum_{i=0}^n {n \choose k} = {n \choose 0} + {n \choose 1} + .. + {n \choose n} ? $
Solution: Using the Binomial Theorem, let x = y = 1. Then, $ \sum_{i=0}^n {n \choose k} (1)^i (1)^{n-i} = \underline{\sum_{i=0}^n {n \choose k}} = (1+1)^n = \underline{2^n}. $
