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==Combinations== | ==Combinations== | ||
| + | ===Claim=== | ||
| + | <math> {n \choose r} = {n \choose r-1} </math> | ||
| + | |||
| + | ===Proof(Counting Arguement)=== | ||
| + | <math> {n \choose r} </math> = # ways to CHOOSE r things from n | ||
| + | = # ways of LEAVING 'n-r' things from n | ||
| + | = # ways of CHOOSING 'n-r' things from n | ||
| + | = <math> {n \choose r-1} </math> | ||
| + | QED | ||
==Binomial Coefficients== | ==Binomial Coefficients== | ||
Revision as of 15:10, 6 September 2008
Contents
Permutations
Combinations
Claim
$ {n \choose r} = {n \choose r-1} $
Proof(Counting Arguement)
$ {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 r-1} $
QED
Binomial Coefficients
Pascal's Triangle
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}. $
