Combinations problem

Take a set having let's say 10 elements, if I make a random selection of its elements (each item selected with 1/2 probability) what is more likely, to have an even or an odd number of items selected? Or maybe they have the same probability?

Obs. Allow all or no items selected.

Try this problem with 5 elements instead of 10.

Odd and even selection have the same probability here, why?

Since any selection that has 3 elements leaves 2 unselected items. So you have the same number of ways to select 2 or 3 items and also the same for 1 and 4, or 0 and 5. Making equal probabilities for even or odd selections.

Look closely and see that this logic would also work if instead of 5 we had 7, 9, or 1997 elements. But it wouldn't work if we had 10 elements!! So back to square one!

Any ideas?

As it turns out odd and even selections also have the same probability in that case! And for any quantity of initial elements for that matter. But why anyway?

Maybe you could use the fact that we already now it is true for for 5 elements!

Indeed! Split the 10 elements in two groups of 5, A and B. How can we make an odd selection of 10 terms?

Case 1: Take an odd number of elements from A and even for B, or
Case 2: Take an even number of elements from A and odd from B.

And for an even number of terms?

Case 3: Take an even number of elements from A and even for B, or
Case 4: Take an odd number of elements from A and odd from B.

Here Case 1 and 3 have the same number of possible selections and Case 2 and 4 also.

So even and odd selection have the same probability in that case too, and also in general following similar lines.