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If $n$ must be greater than or equal to 2, prove that the cardinality of $S$ is a composite number.

Any help would be greatly appreciated.

Edit: I see now that this is more simply a matter of proving that $\binom{2n}{n}$ is always composite. This is explained extremely clearly in this answer.

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  • http://math.stackexchange.com/questions/78533/prove-that-2n-n2-is-even-if-n-is-a-positive-integer – David P Sep 21 '14 at 21:22
  • Can you show that $\binom{2n}{n}$ is always even? – Viktor Vaughn Sep 21 '14 at 21:22
  • I can indeed. :) The question @David Peterson linked to is excellent and gives some very clear examples as to how to solve this. Given that such an excellent explanation exists, I feel it would be best to mark this question as a duplicate. –  Sep 21 '14 at 21:29

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$\binom{2n}{n} = \binom{2n-1}{n} +\binom{2n-1}{n-1}$ and $\binom{2n-1}{n} = \binom{2n-1}{n-1}$, so $\binom{2n}{n}$ is even

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