A set contains $\{1,2,3,4,5....n\}$ where $n$ is a even number. how many subsets that contain only even numbers are there$?$ A set contains $\{1,2,3,4,5....n\}$ where $n$ is a even number. how many subsets that contain only even numbers are there for the set$?$
This is my solution, is this valid$?$
since number of single element subset that contain only a even number is: $n/2$
a element is either in or not in the subset, hence $2$ choices.
Hence $2^{(n/2)}$ would give us all possible combinations of subset that contains only even number, including the empty set. 
Hence my answer is given by $2^{(n/2)} - 1$. subtracting the $1$ because of the empty set $C(n,0)=1$.
 A: Most of your argument is correct, but your answer is wrong. The empty set is a subset that contains only even numbers, so it should not be excluded from the count. In fact, my approach would be to recognise that one is asking for any subset of the set of even numbers from $\{1,2,\ldots,n\}$ (the condition of only being allowed to choose even numbers mean the odd numbers from that set can be ignored), and that set of even numbers has $n/2$ elements, and therefore has $2^{n/2}$ subsets.
A: A set consisting only of even numbers can be constructed in one of the following ways:
$n/2\choose n/2$ ways to construct a set consisting of $n/2$ elements;
$n/2\choose (n/2) -1$ ways to construct a set consisting of $(n/2)-1$ elements;
$n/2\choose (n/2)-2$ ways to construct a set consisting of $(n/2)-2$ elements;
and so on and so forth until
$n/2\choose 1$ ways to construct a set consisting of $1$ element;
$n/2\choose 0$ ways to construct a set consisting of no elements;
Thus the total number of possible subsets is 
$n/2\choose n/2$ + $n/2\choose (n/2)-1$ +...+$n/2\choose 1$+$n/2\choose 0$=$(2)^{n/2}$.
Remark: A set $S$ consisting of only even numbers means that if $x\in S$, then $x$ should be even. This is vacuously true if the set is empty and hence we should include $n/2\choose 0$.
A: Lets prove this through recursion.
Let $a_n$ denotes the number of subsets of the set $S_n =\{1,2, \cdots , n\}$ containing even numbers only. Note that the null set is not a number(don't think it as 0) so we are not counting it.
Note that $a_{2n} = a_{2n+1}$ and $a_2 =1$
Now $$a_{2n+2} = a_{2n} + a_{2n} +1  = 2a_{2n} + 1 $$ Reasoning is based on the fact that the first $a_{2n}$ are the no. of even only subsets in $S_n$ and the second $a_{2n}$ denotes (those subsets only +the element "2n+2" ) is $S_{2n+2}$. The last 1 is for the subset {2n+2} only. 
So, $$a_{2n+2} = 2a_{2n}+1 = 4a_{2n-2} +1 +2  = 2^n \cdot a_2 + \sum_{i=0}^{n-1} 2^i =2^n + 2^n -1 = 2^{n+1} - 1 $$
Hence, $$\boxed{a_{2n} = 2^n -1 }$$
Proved!!! 
