Answer :
The number of ways to choose 10 objects from a set of 31 with 10 identical and 21 distinct is 2 to the power of 21, as each distinct object is either chosen or not. This result is obtained by adding up the combinations of selecting 10 down to 0 distinct objects from the 21 available, which simplifies to 2²¹. Hence the correct answer is option D
The problem at hand requires us to calculate the number of ways to choose 10 objects from a set of 31 objects, where 10 are identical and the remaining 21 are distinct. This is a combinatorial problem that can be solved using combinations and permutations. When choosing from identical objects, the order in which they are selected does not matter, hence it is a combination problem. However, when choosing from distinct objects, the order does matter, making it a permutation problem.
To solve this, we look at the number of ways of choosing 0 to 10 identical objects, and then for each of those ways, we choose the remainder from the 21 distinct objects. Since there is only one way to choose 0, 1, 2, ..., or 10 identical objects, we essentially need to calculate the sum of combinations of choosing 10, 9, 8, ..., 0 distinct objects from the 21 available.
The formula for combinations is C(n, r) = n! / [r!(n - r)!]. So, we calculate the sum C(21, 10) + C(21, 9) + ... + C(21, 0). This results in 221 possibilities, since every distinct object is either chosen or not (which parallels the binary system).
Hence the correct answer is option D