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

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 ?

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

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