The Number of ways of choosing 10 objects out of 31 objects of...

There is only one way in which we can choose one, two or more objects from $10$ identical objects.
Number of ways of selecting $10$ objects
$= {}^{21}{C}_{10} + {}^{21}{C}_{9} + {}^{21}{C}_{8} + {}^{21}{C}_{7} + {}^{21}{C}_{6} + {}^{21}{C}_{5} + {}^{21}{C}_{4} + {}^{21}{C}_{3} + {}^{21}{C}_{2} + {}^{21}{C}_{1} + {}^{21}{C}_{0}$
We know ${}^{n}{C}_{r} = {}^{n}{C}_{n - r}$ and
⇒ {2}^{21} = {}^{21}{C}_{0} + {}^{21}{C}_{1} + {}^{21}{C}_{2} + {}^{21}{C}_{3} + {}^{21}{C}_{4} + . . . . . . . . . + {}^{21}{C}_{21}\left\{$\begin{array}{ll}∵ {}^{21}{C}_{21} = {}^{21}{C}_{0} \\ {}^{21}{C}_{20} = {}^{21}{C}_{1} \\ {}^{21}{C}_{19} = {}^{21}{C}_{2}\end{array}$\right)<br />= 2\left({}^{21}{C}_{0} + {}^{21}{C}_{1} + {}^{21}{C}_{2} + \ldots \ldots \ldots \ldots \ldots + {}^{21}{C}_{10}\right)$<br />$⇒ {}^{21}{C}_{0} + {}^{21}{C}_{1}{+}^{21}{C}_{2} + . . . . . . . . . + {}^{21}{C}_{10} = {2}^{20}