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
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.