# How many ways are there to pick r objects from n objects when each object appears an odd number of times? [duplicate]

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

## marked as duplicate by Mike Earnest, Lord Shark the Unknown, Cesareo, mrtaurho, drhabApr 6 at 9:48

• 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 4 at 19:27
• 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 5 at 5:58
• Alright so just dividing $\frac {n!} {(n-r)! r!}$ by 2 is my final answer? – Brownie Apr 4 at 19:26
• 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 4 at 19:34