The same thing, combinatorially. We want to choose $5$ positive integers from the first $21$. This can be done in $\binom{21}{5}$ ways.
We count the same thing in a different way. If the biggest chosen number is $21$, the rest can be chosen in $\binom{20}{4}$ ways. If the biggest is $20$, the rest can be chosen in $\binom{19}{4}$ ways. If the biggest is $19$, the rest can be chosen in $\binom{18}{4}$ ways. And so on, until if the biggest is $5$, the rest can be chosen in $\binom{4}{4}$ ways. We conclude that $\binom{20}{4}+\binom{19}{4}+\binom{18}{4}+\cdots+\binom{10}{4}+\binom{9}{4}+\cdots +\binom{4}{4}=\binom{21}{5}. \tag{$1$}$ The same reasoning shows that $\binom{9}{4}+\binom{8}{4}+\cdots+\binom{4}{4}=\binom{10}{5}. \tag{$2$}$
Now subtract $(2)$ from $(1)$.