Picking zero out of zero elements

Question

How can this expression$$\binom{0}{0}=1$$ be logical?

I do not know if this is good example:

Probability of selecting one blue candy from a jar, which contains 5 green candies.

$$\frac{\binom{0}{0}}{\binom{5}{1}}=\frac{1}{5}$$

Answer 1

You simply did not set up the probability question right.

The probability of selecting ONE blue candy from five green candies is

$$\frac{0\choose 1}{5\choose 1} = \frac 05 = 0.$$

It isn't $\frac{0\choose 0}{5\choose 1}$ as you wrote.

To get a probability using ${0 \choose 0}$ you should calculate the probability of getting zero of something that doesn't exist: what is the probability of choosing zero blue candies from five green candies.

That'd be $\frac {0 \choose 0}{5 \choose 0}$ which would be $\frac 11 = 1$. You will always choose zero things if the things don't exist.

Colloquially: $n \choose 0$ is the way to choose zero from $n$ and there is always one way to do that: to not choose anything. To choose zero from zero is simply to not pick anything from an empty set. The one way to do that is to not pick anything.

Answer 2

By definition, $0! = 1$. Think of it like this: how many ways can we arrange no objects? Easy - one way, the way that arranges no objects.

Answer 3

There is only one way to choose $0$ objects from any sized set - To not choose. That makes as much sense if the set itself is empty as if the set has multiple objects.

Answer 4

The number of $0$-element subsets of $0$-element set is $1$. Indeed, the only empty subset of $\emptyset$ is $\emptyset$. :-)

Answer 5

Other answers correctly address the first part of your question. For the second, the number of ways to choose one blue candy is $\binom{0}{1} = 0$: the number of ways to choose one item from an empty set. You don't evaluate that with the formula for the binomial coefficients. Use it for the numerator and you get the correct $0$ answer.

Answer 6

First, how many ways are there to choose a subcommittee of $4$ Republicans, $3$ Democrats, and $1$ Libertarian out of a committee of $8$ Republicans, $6$ Democrats, and $2$ Libertarians? The answer is $$ \binom 8 4 \binom 6 3 \binom 2 1. $$ So how many ways are there to choose a subcommittee of $4$ Republicans, $3$ Democrats, and $0$ Libertarians out of a committee of $8$ Republicans, $6$ Democrats, and $0$ Libertarians? This should be $$ \binom 8 4 \binom 6 3 \binom 0 0. $$ But it should also be $$ \binom 8 4 \binom 6 3, $$ since you can just omit all mention of Libertarians in that case.

Answer 7

Note also that $0! = 1$, for good reasons: it's convenient to define the empty product to be the identity, because then we get $$\prod_{x \in A} x \times \prod_{y \in B} y = \prod_{z \in A \cup B} z$$ for disjoint $A, B$ even when $A = \emptyset$.

Answer 8

Definition 1.1 of https://www.math.ucdavis.edu/~hunter/intro_analysis_pdf/ch1.pdf is helpful to understand this.

Consider the power set $\mathcal{P}(X)$ of a set $X$, which is the set of all subsets of $X$. Note that this includes the empty set $\emptyset$. Picking any number (including 0) of items from $X$ is equivalent to picking one of the items in $\mathcal{P}(X)$. Thus, the number of ways of picking any number $n$ of items from $X$ is equal to the number of occurrences in $\mathcal{P}(X)$ of a set of length $n$.

If we select $n=0$, there is only one set of length $0$, namely the empty set $\emptyset$ which exists only once in $\mathcal{P}(X)$. Thus follows ${m \choose 0}=1$ for any $m \in \mathbb{N}$.