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So this is my first time encountering this type of problem, and I'm at quite a beginner level. If you could help me get on the right track I'd appreciate it.

Now if i'm picking r objects from n objects that's going to be $n\choose r$

now there would be $\frac {n!}{(n-r)! r!}$ combinations (i think)

if we just wanted to know ways where each object appears an odd number of times, wouldn't that be half of the total possibilities?

My logic and math may be way off here, but I'm just a beginner. I feel like this makes sense because all the odd possibilities would be half the amount of all possible possibilities right?

So if you were to pick r objects from n objects, how many ways are there to pick so that each r object appears an odd number of times. Is what the question is asking

Brownie
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    It is not clear what are you asking. Are all $n $ objects identical? Or are they all distinct? Both cases are not consistent with the sentence "each object appears an odd number of times". – user Apr 04 '19 at 19:27
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    You have added a sentence which does not help understanding your question. Do you mean "how many ways are there to pick up an odd number of objects out of $n$ distinct items?" What does "r" in "each r object" mean? – user Apr 05 '19 at 05:58
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    I am still confused by what you are asking. Can you give and example? Maybe choose a small enough value of n and r and list out all of the combinations you are talking about? – Mike Earnest Apr 06 '19 at 04:07
  • So if r = 2 and n = 3 and we have {1,2,3} we can pick 2 objects (with repetitions) from the n (3) objects. (1,1) (1,2) (1,3) (2,2) (2,1) (2,3) .. and so on then the question is asking how many of those "groups" does each object appear an odd number of times. This is at least how I understood the question, and is possible quite wrong. Especially after seeing some of the responses, and linked duplicate I am more confused. I will get some clarification and update! Thanks for the help – Brownie Apr 07 '19 at 02:13

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Yes your intuition is correct. Observing the pascals triangle will give you a proof of this. I assume your familiar with the connection between the triangle and the binomial coefficients. Each number in the triangle is the sum of the 2 above it so look at each number in a row. It adds it's value to two adjacent numbers in the row below so it contributes the same amount to both the sum of the odd numbered entries and the sum of the even numbered entries.

G Aker
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  • Alright so just dividing $\frac {n!} {(n-r)! r!}$ by 2 is my final answer? – Brownie Apr 04 '19 at 19:26
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    No, your formula gives you how many ways you can choose r object from n. So you need to sum over all odd number values for r. You could also sum over every choice of r which would give you $2^n$ and devide by 2 for $2^{n-1}$. – G Aker Apr 04 '19 at 19:34