TITLE: How many subsets of set $\{1,2,\ldots,10\}$ contain at least one odd integer?
QUESTION [16 upvotes]: How many subsets of set $\{1,2,\ldots,10\}$ contain at least one odd integer?     

My Working:
What I can think of is subtracting the no. of subsets that don't contain a single odd number from the total number of subsets as if we calculate it for single cases (like $1$ odd integer, $2$ odd integers, $\ldots$) it would be pretty long.    
As there needs to be no odd integer, the maximum number of elements in the set is $5$ (only $5$ even integers are there in the superset).   
Case 1: $0$ elements: $1$ set
Case 2: $1$ element: $5$ sets ($1$ even integer in each set)
Case 3: $2$ elements: $(5)(5)$ sets ($1$ element odd and $1$ even) $+\binom{5}{2}$  sets (both elements even)
which gives $35$ sets     
Case 4: $3$ elements: $\cdots$
Problem:
This is getting complicated and I'm pretty sure I'll mess up if I proceed further. Is there any other way of solving this question?

REPLY [6 votes]: A subset of $\{1,2,\ldots,10\}$ that contains at least one odd number is of the form $A \cup B$, where $A$ is a subset of $\{2,4,6,8,10\}$ and $B$ is a non-empty subset of $\{1,3,5,7,9\}$. The set $\{2,4,6,8,10\}$ has $2^5=32$ subsets. The set $\{1,3,5,7,9\}$ has $2^5=32$ subsets, of which $31$ are nonempty. Therefore, the answer is $32 \cdot 31 = 992$.