Why is the number of ways of choosing $0$ items from $n$ items $1$?

Solution 1:

When you phrase it in plain English, the answer isn't necessarily totally clear. It might seem reasonable to argue that there are $0$ ways of choosing $0$ things from $n$, since choosing $0$ things isn't really a choice. However, when mathematicians talk about "the number of ways to choose $0$ from $n$," we mean something a bit more specific.

A few ways of describing what we mean:

1. The number of subsets of an $n$-element set with $0$ elements (and we always assume the empty set counts)
2. The number of functions from a $0$-element set to an $n$-element set, up to permutations of the domain (there's exactly one function from the empty set to any set).
3. The coefficient of $x^0y^n$ in the expansion of the binomial $(x+y)^n$.

All of these agree that there is one way to choose $0$ out of $n$ things, and all of these perspectives are mathematically very useful. Thus the perspective that there is a way to choose $0$ out of $n$ things is universal in mathematics.

This wasn't always so obvious to everyone: for instance, I have heard of an early 20th century mathematician who insisted in his books that all intersections be nonempty to be defined. It seems to me that this would be consistent with refusing to accept the empty set as a subset, and with saying there are $0$ ways to choose $0$ things from $n$. But this would make for lots of inconvenient circumlocutions, so we've settled pretty firmly on the other solution.

Solution 2:

Perhaps a little duality argument can help to provide some intuition.

Given a bag of $n$ balls, consider these two tasks:

1. choose $k$ balls inside the bag, and remove them from the bag
2. choose $n-k$ balls inside the bag, and remove all the others from the bag

Intuition suggests that these tasks are essentially the same: choosing which $k$ balls we remove corresponds to choosing which $n-k$ balls we keep in the bag.

Indeed, there are as many ways to choose $k$ balls (to remove) as there are to choose $n-k$ balls (to keep).

In particular, to determine how many ways we have to choose (and remove) $0$ balls, we can equivalently count how many ways we have to choose (and keep) $n-0=n$ balls. In your own question, you agree on there being only one way to choose $n$ balls out of $n$. Hence, "one way to choose what to keep" can be restated as "one way to choose what to remove".