Power sets question

[perusal]

Can someone explain to me why P({1,2,3}) is equal to 2^3

I understand how 8 is arrived at. {empty set, {1},{2},{3},{2,3},{1,2},{1,3},{1,2,3}}

What I don't understand is why it how this relates to 2^3

Has it got something to do with the binomial theorem?

 

[MarkFL]

Yes, we can state:

[MATH]P_n=\sum_{k=0}^n\left({n \choose k}\right)=\sum_{k=0}^n\left({n \choose k}1^{n-k}\cdot1^{k}\right)=(1+1)^n=2^n[/MATH]

 

[HallsofIvy]

It relates to the fact that 2^3 is equal to 8!

 

[perusal]

Yes, we can state:

[MATH]P_n=\sum_{k=0}^n\left({n \choose k}\right)=\sum_{k=0}^n\left({n \choose k}1^{n-k}\cdot1^{k}\right)=(1+1)^n=2^n[/MATH]

Thanks for your help.

How do we get from

[MATH]\sum_{k=0}^n\left({n \choose k}1^{n-k}\cdot1^{k}\right)[/MATH]
to

[MATH](1+1)^n[/MATH]

 

[MarkFL]

Thanks for your help.

How do we get from

[MATH]\sum_{k=0}^n\left({n \choose k}1^{n-k}\cdot1^{k}\right)[/MATH]
to

[MATH](1+1)^n[/MATH]

Via the binomial theorem.

 

[Dr.Peterson]

Think about how you expand (a + b)^n by the binomial theorem. Then replace a and b with 1.

 

[pka]

How do we get from [MATH]\sum_{k=0}^n\left({n \choose k}1^{n-k}\cdot1^{k}\right)[/MATH]to [MATH](1+1)^n[/MATH]

You need to know the basic binomial expansion formula:
$\displaystyle {(x + y)^n} = \sum\limits_{k = 0}^n {\binom{n}{k}{x^{n - k}}{y^k}} = {x^n} + n{x^{n - 1}}y + \cdots nx{y^{n - 1}} + {y^n}$
Now if $\displaystyle x=1~\&~y=1$ we have $\displaystyle (1+1)^n=2^n$
Thus $\displaystyle {(2)^n} = \sum\limits_{k = 0}^n {\binom{n}{k}}$
All you need to know now $\displaystyle \binom{n}{k}=\frac{n!}{k!(n-k)!}$ from $\displaystyle n$ choose $\displaystyle k$, a subset of size $\displaystyle k$ in a set of size $\displaystyle n$.

 

[perusal]

Thanks for all the help. I understand now