Question: Let $X=\{1, 2, …, n\}$. where $n$ is a positive integer. Let $n$ be odd. How many subsets of $X$ contain all the odd numbers from $X$?
Answer: Since $n$ is odd, $n=2k+1$ for some integer $k\ge0$, so there are $k$ even numbers and $k+1$ odd numbers in $X$. Thus, there are $2^k=2^{(n-1)/2}$ subsets of $X$ containing all the odd numbers in $X$.
I'm very confused by the answer.
If there are $k+1$ odd numbers in $X$, why are there then $2^k=2^{(n-1)/2}$ subsets that only contain odd numbers?
From my understanding, we want to have all subsets of $X=\{1, 2, …, n\}$ that contain only odd numbers. I thought, when $n$ is odd, there are $(n-1)/2$ even numbers and $(n+1)/2$ odd numbers in the set.
But the answer gives a different value. Or do I misunderstand the wording here?
Best Answer
The question is asking about subsets that contain all the odd numbers, that does not necessarily mean ONLY odd numbers. E.g., if $n=5$, then $\{1, 2, 3, 5\}$ contains all odd numbers.