Number of ways of selecting zero items out of $n$ objects = 1. But why? There are $n$ things (all distinct). You have to select zero thing (or things ?) out of it. Number of ways of doing this task is $1$.
But why is that ?
I asked to my teacher. They said it is because number of ways of selecting $r$ things out of $n$ things is equal to the number of ways of un-selecting $(n-r)$ things out of those $n$ things. Hence, selecting zero thing here is same as not selecting exactly all things, which can be done in 1 way.
But if I ask that in how many ways you can un-select exactly zero things out of $n$ things? Giving same explanation as above will not work. Because this explanation is circular. It's really not a proof.
On internet, I got to know that it's because $^nC_0 = 1$. But then question reduces to prove that this formula of $^nC_r$ must work for $r=0$ ( I am not asking to prove $0! = 1)$ i.e. $^nC_0$ represents the number of ways to do it, as it works for every other positive integer.
Please shed some light on it. This question is troubling me from many days.
EDIT:
If I understand the answers given below perfectly, then it is because empty sets correspond to that $1$ way. And all empty sets are identical because they are subsets of each other. They are subsets of each other because it's their definition i.e. it's an axiom.
 A: The empty set is a set.  It is just as valid of a mathematical object as any other, and one might say even more fundamental of a mathematical object than most since its existence is generally one of the founding axioms of whatever set theory you are working with.
The binomial coefficient $\binom{n}{k}$ is traditionally best worded in terms of pure set theory.  It is defined to be the number of sets of size $k$ who are subsets of the set $[n]$, the prototypical set of $n$ elements

(whether you prefer that to be $\{1,2,3,4,\dots,n\}$ or if you prefer that to be $\{0,1,2,3,\dots,n-1\}$ is irrelevant.  Note also that $[0]=\emptyset$ and $[1]$ is equal to $\{1\}$ or $\{0\}$ depending on preference)

Since there is in fact one set who is of size zero who is a subset of your set $[n]$ it follows that $\binom{n}{0}=1$... that set, namely, the empty set.
A: Since the other answer goes the more theoretical route, I'll try the applied route.
You are in a section meeting at work with your $n-1$ colleagues (which makes you $n$ people in total), and you need to select a committee to handle, say, the watering of plants in the office. Instead of actually paying attention to the meeting, you decide to take a block of post-it notes and write one possible committee configuration on each of them, until you have listed them all.
How many of your post-it notes have a two-person committee on them? It's $^nC_2$, and some reasoning (or knowing it already) will lead you to the fact that there are $\frac{n(n-1)}2$ such notes, so $^nC_2 = \frac{n(n-1)}2$. How many have a single-person committee on them? It's $^nC_1$, and also it's $n$, so we must have $^nC_1=n$. How many notes are blank? Well, it's $^nC_0$, and also clearly it is $1$.
On the other hand, there are $0$ notes with $n+1$ names on them, because that would be impossible to do, and this tells us that $^nC_{n+1} = 0$.
A: There is exactly $1$ way to do nothing.
A: You must realise first there is a difference between having one option and none at all; even though in the first case we informally say one has no choice, there is still a way to satisfy the requirements. If there are $3$ objects and you must select $3$ of them, then you have no choice (within the requirements) but to select each one of them, but that is still an option. If there are $3$ objects and you must select $4$ (or more) of them then there is no option at all that satisfies the requirement. Similarly if you must select $0$ of the object there is no choice but to select none, but it can still be done satisfying the requirement; if you had been asked to select $-1$ of them that's simply no possible at all.
A: Suppose there are 10 players in the room who are the part of a game team. According to rules, there should be 11 players to complete the team and initiate the game. Two candidates (say A and B) applied for joining in the team. How many distinct teams can be made ?
The answer is two. The two teams are :

*

*ten previous players + A


*ten previous players + B
Now suppose that the rules are changed such that 12 players are required to complete the team and initiate the game. Now, how many distinct teams can be made ?
The answer is one. There's only one way i.e. add both A and B to team.
Now suppose that the rules are changed again so that only 10 players are required to complete the team. How many teams can be made so as to initiate the game ?
Obviously the answer is one. By both rejecting A and B, our team is complete. Hence, a unique game could be played without selecting any of A or B. Thus, number of ways to do nothing is 1.
