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?
Best Answer
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.