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How many combinations of size $n+1$ at least in a group of $2n$ elements ?

It seems that total amount of combinations is equal to $2^{2n}$ in this case, but what if I only want the number of combinations strictly bigger than n elements ?

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    $\begingroup$ See Pascal's Triangle and examine (for example) $$\sum_{k=3}^4 \binom{4}{k}, ~\sum_{k=4}^6 \binom{6}{k}, ~\sum_{k=5}^8 \binom{8}{k}, \cdots .$$ $\endgroup$ Commented Feb 3, 2021 at 14:36

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Slightly poorly phrased question, but a pretty simple answer. We know $$ {a\choose b} = {a\choose a-b} $$

and so we have that $$ {2n\choose x} = {2n\choose 2n-x} $$

This gives us a nice symmetry in your 2n elements between the combinations of size less than $n$ and greater than $n$. Thus if we take $2^{2n} - {2n\choose n}$ we have exactly twice the number of combinations of size greater than $n$. This gives us our final answer as $$ \frac{2^{2n} - {2n\choose n}}{2} $$

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You would have to subtract $\binom{2n}{n}$ from $2^{2n}$ and then divide by $2$.

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    $\begingroup$ why would you divide by 2? $\endgroup$
    – Jewgah
    Commented Feb 3, 2021 at 14:36
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    $\begingroup$ @TheUsualProcess see my comment, which follows your query. $\endgroup$ Commented Feb 3, 2021 at 14:37

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