In a set A ={1,2,3,4,5,7,8,10,11,14,17,18} how many subsets of A contain only Odd numbers? I know the answer is 64 from 2^6 but I may just not be great at logic or fully know the definition of a power because I am confused on the reasoning why.
The reasoning I heard is you have 2 choices: to include an odd number or not and since you have 6 odd numbers the answer is 2^6. 
I am just a bit lost on how the power represents all the subsets if there is a another way to do this perhaps with combination formulas i'd also appreciate it. (I am in Intro to Discrete)
Edit: For clarity, I mean how many subsets of A are there that contains only odd numbers. So I assume {1,3} , {1,3,5} {3,5,1} etc in that sense Also changed the title it was a typo with I mean only not all.
 A: The question is really "how many subsets of $\{1,3,5,7,11,17\}$ are there, not counting the null set?"
As you said each element is either in the subset or it isn't. Thus each element has two choices, so there are $2\cdot2\cdot2\cdot2\cdot2\cdot2=2^6$ subsets, and we subtract one more since we also just counted the null set, which we don't want.
This may be easier to see with two elements, $\{a,b\}$. If $1$ represents "in the subset" and $0$ represents "not in the subset" then all possible subsets are
$0,0$, or $\emptyset$
$0,1$, or $\{b\}$
$1,0$, or $\{a\}$
$1,1$, or $\{a,b\}$
A: Let $p =$ number of elements of a set $S$
Maximum number of subsets of $S$ is $2^p$ which includes a null set too. 
Since, we want only odd numbers so we take a set of odd numbers
O ={1,3,5,7,11,17} which has cardinality $6$. 
Then the maximum number of sets that can be formed from this set is $2^6 -1=63$ because we have to discard the null set. 
A: 
The reasoning I heard is you have 2 choices: to include an < element > or not.

Yes, that is the reason.  Let's see if we can make it click with you.
Let's avoid extraneous distractions.  How many ways are there to choose from $\{A,B,C\}$.  They answer should be $2^3 = 8$.  But why?  Each element may be chosen or not.  Let's label the ones we choose with $Y$ and the ones we ignore with $N$.
Here are our $8$ choices:
$$\begin{align}
A_y\ B_y\ C_y &\to \{A,B,C\}\\
A_y\ B_y\ C_n &\to \{A, B\}\\
A_y\ B_n\ C_y &\to \{A, C\}\\
A_y\ B_n\ C_n &\to \{A\}\\
A_n\ B_y\ C_y &\to \{B,C\}\\
A_n\ B_y\ C_n &\to \{B\}\\
A_n\ B_n\ C_y &\to \{C\}\\
A_n\ B_n\ C_n &\to \{\} \to \text{empty set}
\end{align}$$
Does that make it click?
