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Question based on Permutations and Combinations

The number of ways of selecting 10 objects from 30 objects, of which 10 are alike and the rest are all different is ?

A) $2^{20} + binom{20}{10}$

B) $2^{20} - binom{20}{9}$

C) $2^{19} + binom{20}{9}$

D) $2^{19} - binom{20}{10}$

If you choose $n$ of the unique object, then there are $20choose n$ ways to pick those, and $1$ to pick the other $10-n$ non-unique ones. So the answer is
$$sum_{n=0}^{10} {20choose n}$$

By symmetry, we have
$$2sum_{n=0}^{10} {20choose n} = sum_{n=0}^{20} {20choose n} + {20choose 10} = 2^{20} + {20choose 10}$$

So
$$sum_{n=0}^{10} {20choose n} = 2^{19} + frac12{20choose 10}$$

And ${19choose10} = {19choose9}$ and ${19choose10} + {19choose9} = {20choose10}$ by Pascal's Identity, so this gives $frac12 {20choose10} = {19choose9}$ so the final answer is
$$2^{19} + {19choose 9}$$

Suppose you pick $p$ objects from the set $P$ of identical objects, and let the set $Q$ be that of the other $20$ objects. Scaling $p$ in the range $[0, 10]$ and permuting the ways to complete the $10$ with $Q$ gives us:

$$sum_{p=0}^{10}binom{20}{10-p}$$