How many subsets of set $\{1,2,\ldots,10\}$ contain at least one odd integer?

This is a classic case where looking at the excluded space is far easier.

Any subset without at least one odd integer is a subset of $\{2,4,6,8,10\}$.

There are $2^{10}$ subsets of $\{1,2,3,4,5,6,7,8,9,10\}$ and $2^5$ subsets of $\{2,4,6,8,10\}$. So there are $2^{10}-2^5$ $=1024-32$ $=992$ subsets of $\{1,2,3,4,5,6,7,8,9,10\}$ which include at least one odd number.

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:

$2^5 \sum_{k=1}^5 \binom{5}{k} = 2^5 \left( \sum_{k=0}^5 \binom{5}{k} - \binom{5}{0} \right) = 2^5 \left( 2^5 - 1 \right) = 2^{10} - 2^5 = 992$

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 $32 \cdot 31 = 992$.