# Power sets question

#### perusal

##### New member
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

##### Super Moderator
Staff member
Yes, we can state:

$$\displaystyle P_n=\sum_{k=0}^n\left({n \choose k}\right)=\sum_{k=0}^n\left({n \choose k}1^{n-k}\cdot1^{k}\right)=(1+1)^n=2^n$$

#### HallsofIvy

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

#### perusal

##### New member
Yes, we can state:

$$\displaystyle P_n=\sum_{k=0}^n\left({n \choose k}\right)=\sum_{k=0}^n\left({n \choose k}1^{n-k}\cdot1^{k}\right)=(1+1)^n=2^n$$

How do we get from

$$\displaystyle \sum_{k=0}^n\left({n \choose k}1^{n-k}\cdot1^{k}\right)$$

to

$$\displaystyle (1+1)^n$$

#### MarkFL

##### Super Moderator
Staff member

How do we get from

$$\displaystyle \sum_{k=0}^n\left({n \choose k}1^{n-k}\cdot1^{k}\right)$$

to

$$\displaystyle (1+1)^n$$
Via the binomial theorem.

#### Dr.Peterson

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

#### pka

##### Elite Member
How do we get from $$\displaystyle \sum_{k=0}^n\left({n \choose k}1^{n-k}\cdot1^{k}\right)$$
to $$\displaystyle (1+1)^n$$
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

##### New member
Thanks for all the help. I understand now