Expected value of the largest element of a subset

There are no subsets of $10$ elements with $1,2,3...,9$ as the maximum element. The number of subsets with $m(\ge10)$ as the maximum element is given by $\binom{m-1}9$, since you select $9$ other numbers from the $m-1$ numbers smaller than $m$. There are $\binom{20}{10}$ subsets of size $10$ in totality. The required expectation is$$\frac1{\binom{20}{10}}\sum_{m=10}^{20}m\binom{m-1}9=\frac1{9!\cdot\binom{20}{10}}\sum_{m=10}^{20}\frac{m!}{(m-10)!}$$Now use the fact

$\frac{m!}{(m-10)!}=m(m-1)...(m-9)\\=\frac{(m+1)-(m-10)}{11}[m(m-1)...(m-9)]\\=\frac1{11}[(m+1)m...(m-9)-m(m-1)...(m-10)]$

and so$$\sum_{m=10}^{20}\frac{m!}{(m-10)!}=\frac{21\times20\times19\times...\times11}{11}=21\times20\times19\times...\times12$$giving the required expectation as $210/11$.

In general, when subsets of size $q$ of the set $\{1,2,...,n\}~(q\le n)$ are taken and their maximum noted, the expected maximum is$$\frac{(n+1)q}{q+1}$$

Take $10$ red balls and $11$ green balls and uniformly randomly arrange them in a circle. By symmetry, each of the $10$ red balls has equal probability $\frac1{11}$ of being in any of the $11$ intervals between pairs of green balls. Uniformly randomly pick one of the green balls and remove it to split the circle into a line, leaving $10$ green balls and $10$ red balls. Number them from $1$ to $20$ along the line. The remaining $10$ green balls represent the $10$ numbers you pick. Each red ball has probability $\frac1{11}$ of being above the highest green ball. Thus we expect $\frac{10}{11}$ red balls to be above the highest green ball, and thus we expect its number to be $20-\frac{10}{11}=\frac{210}{11}$.