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I have a set of elements $\mathcal{P}$ of dimension $2N$. I imagine $\mathcal{P}$ as a collection of pairs, e.g. for $N=2$

$$ \mathcal{P}:=\{\underbrace{a_1,a_2}_{\text{pair}}, b_1, b_2\} $$

I want to construct all the possible unordered collections of $k\leq 2N$ elements from $\mathcal{P}$ such that

(1) there are no repetitions

(2) if an element of a pair is included, the other is excluded

Continuing the example above: for $k=2$ the collections are $$ \{a_1,b_1\}, \{a_1,b_2\}, \{a_2,b_1\}, \{a_2,b_2\} $$ for $k=3,4$ there are no collections

for $k=1$ the collections are $$ \{a_1\}, \{a_2\}, \{b_1\}, \{b_2\} $$ How many collections are there for any generic $N$ and $k$?

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    I'm confused., what about $\{a_1\},\{a_2\},\{b_1\},\{b_2\}$ and $\emptyset$. Are these not also sets which satisfy the conditions of "no repititions" and "if an element of a pair is included the other is excluded"? You didn't specify that $k$ must be precisely two in your example. If that was the case that you forgot to specify $k=2$ then would $k=1$ be the four nonempty sets I pointed out at the beginning? Or are you requiring one of every pair be used? – 2017-02-20
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    Regardless... if you allow any value of $k$... pick which $k$ "letters" are used, then for each "letter" pick which "number" is used for it. It is ambiguous whether $N$ is the number of pairs or the number of elements since you seem to have gone back and forth between the two... letting $N$ be the number of pairs, you would have $\binom{N}{k}2^k$ collections of size $k$ – 2017-02-20
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    Thanks, I have edited my question to clarify the points you pointed out. – 2017-02-20

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