Understand the objects

We have total $31$ objects made up of:

• $10$ identical objects of one kind
• $21$ distinct objects

We need the number of ways to choose $10$ objects.

Case-wise counting

Since the $10$ identical objects are all the same, if we choose $k$ of them, that contributes only one way for that part, where $0 \le k \le 10$.

Then the remaining $10-k$ objects must be chosen from the $21$ distinct objects, which can be done in $\binom{21}{10-k}$ ways.

Hence total number of selections is $\sum_{k=0}^{10} \binom{21}{10-k}$

Reindexing with $r=10-k$, this becomes $\sum_{r=0}^{10} \binom{21}{r}$

Use symmetry of binomial coefficients

For $n=21$, $\sum_{r=0}^{21} \binom{21}{r} = 2^{21}$

Since $21$ is odd, the terms split symmetrically as $\sum_{r=0}^{10} \binom{21}{r} = \sum_{r=11}^{21} \binom{21}{r} = 2^{20}$

Therefore the required number is $2^{20}$

Match with options

The options appear to be printed as $2^{20}-1$, $2^{20}$, $2^{20}+1$, and $2^{21}$ (written as 220, 221 etc. in plain text).

So the correct option is: $\boxed{2^{20}}$

which is Option B.