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?

Question

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.

Answer 1

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 $\{a,b\}$

Answer 2

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.

Answer 3

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?