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We have <math>\binom{n}{0}+ \binom{n}{1}+...+\binom{n}{n}=(1+1)^n=2^n</math> | We have <math>\binom{n}{0}+ \binom{n}{1}+...+\binom{n}{n}=(1+1)^n=2^n</math> | ||
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| + | == Using Induction == | ||
| + | |||
| + | '''Base case:''' <br> | ||
| + | n=0: <math>2^0=1</math> Subsets with 0 elements: {∅} <br> | ||
| + | n=1: <math>2^1=2</math> Subsets with 1 elements: {∅}, {1} | ||
| + | |||
| + | So we can assume a set ''S'' with ''n'' elements has ''2^n'' subsets. | ||
| + | |||
| + | n+1: <math>2^(n+1) = 2^1 + 2^n = 2*2^n = 2^(n+1)</math> | ||
| + | |||
| + | -Jesse Straeter | ||
| + | |||
| + | ---- | ||
Revision as of 10:45, 7 September 2008
Using Binomial Theorem, $ (a+b)^n=\binom{n}{0}a^n+ \binom n 1 a^{n-1} b+...+\binom{n}{n}b^n $.
We have $ \binom{n}{0}+ \binom{n}{1}+...+\binom{n}{n}=(1+1)^n=2^n $
Using Induction
Base case:
n=0: $ 2^0=1 $ Subsets with 0 elements: {∅}
n=1: $ 2^1=2 $ Subsets with 1 elements: {∅}, {1}
So we can assume a set S with n elements has 2^n subsets.
n+1: $ 2^(n+1) = 2^1 + 2^n = 2*2^n = 2^(n+1) $
-Jesse Straeter
