<s>[INST] How many subsets of set $\{1,2,\ldots,10\}$ contain at least one odd integer? 
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?
 [/INST] Others have already explained the easy solution, here is an alternative more similar to what you tried.
We want to know how many subsets contain exactly $k$ odd integers, from $k = 1$ to $5$, and sum.


*

*$k = 1$: $\binom{5}{1} = 5$ possible subsets of $\{1,3,5,7,9\}$

*$k = 2$: $\binom{5}{2} = 10$ possible subsets of $\{1,3,5,7,9\}$

*$k = 3$: $\binom{5}{3} = 10$ possible subsets of $\{1,3,5,7,9\}$

*$k = 4$: $\binom{5}{4} = 5$ possible subsets of $\{1,3,5,7,9\}$

*$k = 5$: $\binom{5}{5} = 1$ possible subsets of $\{1,3,5,7,9\}$


In each case, we can add some even integers, so we multiply by $2^5 = 32$.
Then,
$2^5 \sum_{k=1}^5 \binom{5}{k} = 32 (5+10+10+5+1) = 992$
But effectively this can be simpler:
$2^5 \sum_{k=1}^5 \binom{5}{k} = 2^5 \left( \sum_{k=0}^5 \binom{5}{k} - \binom{5}{0} \right) = 2^5 \left( 2^5 - 1 \right) = 2^{10} - 2^5 = 992$
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