How many subsets of the set $\{1, 2, \ldots, 11\}$ have median 6?
So I have split this problem into cases. The first case is if 6 is in the subset and the second is where 6 is not.
In case 1, I did 6 with 0, 1, 2, 3, 4, and 5 numbers surrounding it which yielded 1+16+36+16+1 = 70
My struggle is with case 2, how do I find all the unique subsets that don't use 6?
On
As far as the sets that DO contain $6$, we have the subset $\{6\}$, and this is the only way with one element in the subset.
Next we have $\{a,6,b\}$ and there are $5$ choices for $a$ and $5$ choices for $b$, giving a total of $25$ ways with three elements.
Following suite, we have $\{a,b,6,c,d\}$ A little more complicated, but we have $10$ ways for the left side of six and $10$ ways for the right, giving $100$ more subsets.
$\{a,b,c,6,d,e,f\}$ has $10$ ways on each side, giving us another $100$ subsets.
$\{a,b,c,d,6,e,f,g,h\}$ has $5$ ways to arrange each side, so there's $25$ more subsets.
The next size up gives us the original set, so that will be $1$ more
So far we have a total of $\color{red}{252}$ subsets.
Use the tip that Misha provided in their answer to now include the rest of the subsets that do not include $6$ to arrive at the final answer
For a set not containing $6$ to have a median of $6$, it must have an even number of elements, the middle two of which average to $6$. So they can be $5$ and $7$, or $4$ and $8$, or $3$ and $9$, or $2$ and $10$, or $1$ and $11$.
Each of these is handled identically to the case where $6$ occurs in the set and is the median. For example, in the "$4$ and $8$" case, we may add any number of elements from $\{1,2,3\}$ to our set, provided we add the same number of elements from $\{9,10,11\}$, so the number of sets that fall under this case is $\binom30^2 + \binom31^2 + \binom32^2 + \binom33^2 = 1 + 9 + 9 + 1 = 20.$
By the way, you should double-check your arithmetic for your first case. Your approach is right, but you should get $1 + 25 + 100 + 100 + 25 + 1 = 252$ instead of $70$.