counting elements out of the empty set The question is in how many ways can we select 20 different items from the empty set?
ans:
Obviously in 0 ways since the empty set has no items. I mean, this seem obvious, but maybe there is a trick to this question.
 A: The answer is pretty clear if you formulate this question slightly more precisely:
How many sets are there of size $20$ are there that are subsets of the empty set?
We have $\{A| A \subseteq \emptyset \text{ and } |A| = 20\} = \emptyset$ since $A \subseteq \emptyset \implies A = \emptyset \implies |A| = 0$.  So the answer is $0$.
A: If you wanted to sound more mathematical you could make an induction argument:
There are exactly zero ways to select $n>0$ items from the empty set.
Proof: Take $n =1$. Since the empty set has no members it is impossible to select a member of the empty set, thus zero ways to select a single member. Assuming that there are zero ways to select $n$ members, there also must be zero ways to select $n+1$ members, for if there were more than zero ways to select $n+1$ members but zero ways to select $n$ members, it would imply there are more than zero ways to select a single member, contradicting the base case.
It doesn't change the answer but the style is somewhat more formal...
A: One second ago there was a very interesting solution here: $$\binom{x}{k}=\frac{x(x-1)\cdot\ldots\cdot(x-k+1)}{k!}$$ for $x\in\Bbb R$ and $k\in\Bbb N$. The number of choices of $k$ distinct elements from $n$-element set is $\binom{n}{k}$ and $\binom{0}{20}=0$. I like this solution, but I am not the author. If it will return here, I will delete my answer.
