Philippine Mathematical Olympiad, National Orals

Solution:

If $a$ is the smallest element of such a set, then $11 - a$ is the largest element, and for the remaining elements we may choose any (or none) of the $10 - 2a$ elements $a + 1, a + 2, \ldots, (11 - a) - 1$. Thus there are $2^{10 - 2a}$ such sets whose smallest element is $a$. We also require that $11 - a \geq a$ or $a < 6$. We see that there are
$$\sum_{a=1}^{5} 2^{10 - 2a} = \sum_{a=0}^{4} 4^{a} = \frac{4^{5} - 1}{4 - 1} = \frac{1023}{3} = 341$$
possible sets.