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

Question

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

Answer 1

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

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

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

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

Answer 2

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:

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