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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?

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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.

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    $\begingroup$ Yes you are right, I'm not a native speaker and somehow got confused by the wording. So subsets that contain all odd numbers can also contain $0$ or more even numbers. $\endgroup$ – Max Mar 2 at 5:12
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I think you get confused with the explanation of the question already. See again, the question is asking those subsets which contain all the odd numbers, and this doesn't mean that it should contain only odd numbers. And I just noticed that EuxhenH had already give the similar explanation too haha. Hope it helps !

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