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What proportion of subsets of the digits 0-9 contain just as many odd numbers as even numbers? You need to determine how many subsets there are. Then you need to recall that you only need to consider subsets of each even size.

Self attempt: So subset can be represented by $2^x$. We have $2^{10}$ subsets for 0-9. So for odd integers, we have 5 of them (1,3,5,7,9) so $2^5$ and for even integers, we have $2^5$ subsets. Is this correct thinking?

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It’s true that there are $2^5$ sets of odd digits and $2^5$ sets of even digits (including in both cases the empty set of digits), but this is not especially helpful. I suggest that you consider separately sets containing $0,1,2,3,4$, and $5$ odd digits. For instance, there are $\binom53$ sets of exactly $3$ odd digits. To form a set with $3$ odd and $3$ even digits, you must pair this set with any of the $\binom53$ sets of $3$ even digits. There are $\binom53^2$ such pairs (why?), so there are $\binom53^2$ sets of $3$ odd and $3$ even digits.

Now do this for each possible size, and add up the results. If you’ve learned Vandermonde’s identity, you’ll be able to express the answer in an especially nice form, but in any case the numbers are small enough that you can get a numerical answer.

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There are $2^{10}$ subsets.

Call a subset good if it has just as many even as odd.

Let us count how many $2k$-element good subsets there are. The $k$ evens can be chosen in $\binom{5}{k}$ ways, and for each such way the $k$ odds can be chosen in $\binom{5}{k}$ ways, for a total of $\binom{5}{k}\binom{5}{k}$.

To count all good subsets, add up from $k=0$ to $k=5$.

The calculation is not hard, and there are even shortcuts, since $\binom{5}{i}=\binom{5}{5-i}$.

The number of good subsets turns out to be $1^2+5^2+10^2+10^2+5^2+1^2$.

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If you are considering subset will have only even or odd, its incorrect. Some of subset will have both even and odd integer. Also $ \phi $ is common subset in both argued by you so no. of even integers subset and odd integer subset will be $ 2^5 - 1 $ each.

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