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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.

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

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    $\begingroup$ How many distinct ways are there to pick nothing out of nothing? $\endgroup$ – Mehrdad Jan 18 '17 at 8:49
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    $\begingroup$ Because a mathematician said so. That article addresses the question of what 0^0 is but it's the same idea: we define (0 choose 0) in the way that means we don't have to add special clauses for edge cases to our usual formulae. $\endgroup$ – Bristol Jan 18 '17 at 10:42
  • $\begingroup$ @Bristol Thanks for that link. $\endgroup$ – StubbornAtom Jan 26 '17 at 9:43
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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.

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    $\begingroup$ Your last paragraph is not complete unless you add a discussion of what "distinct ways to choose" means -- why is there only one way "not to pick anything", and not more? Because two ways of picking elements are defined to be different if there is an element such that... $\endgroup$ – Federico Poloni Jan 18 '17 at 13:15
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    $\begingroup$ Reading this answer made me remember, ask Siri "what's zero divided by zero?" Someone put a thought into that answer :) $\endgroup$ – nurchi Jan 18 '17 at 21:09
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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.

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    $\begingroup$ As is the usual practice. $\endgroup$ – hobbs Jan 18 '17 at 4:46
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    $\begingroup$ Now american politics invades even math. $\endgroup$ – Tomáš Zato Jan 18 '17 at 14:08
  • $\begingroup$ @TomášZato "Now"? I think you're out of date. I once taught from a book that explicitly said in the introduction that set theory was omitted for religious reasons! And in America, religion and politics are surprisingly mixed despite our first constitutional amendment nominally preventing such mixture. $\endgroup$ – Todd Wilcox Jan 19 '17 at 7:33
  • $\begingroup$ @ToddWilcox: Maybe that was in response to Cantor's religion of set theory? $\endgroup$ – user21820 Jan 19 '17 at 11:39
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    $\begingroup$ @user21820 That's what they said. If I could I would ask them about God's decision to create Kurt Gödel. $\endgroup$ – Todd Wilcox Jan 19 '17 at 14:00
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The number of $0$-element subsets of $0$-element set is $1$. Indeed, the only empty subset of $\emptyset$ is $\emptyset$. :-)

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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.

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  • $\begingroup$ See my comment to fleablood's answer - it applies to yours, too. $\endgroup$ – Federico Poloni Jan 18 '17 at 13:15
  • $\begingroup$ Um "to not choose" $\ne$ "to choose nothing". If you actually "not choose", then you're not choosing, so it's not a way of choosing $0$ objects from the set. $\endgroup$ – user21820 Jan 19 '17 at 11:41
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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.

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  • $\begingroup$ See my comment to fleablood's answer - it applies to yours, too. $\endgroup$ – Federico Poloni Jan 18 '17 at 13:16
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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 $A = \emptyset$.

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    $\begingroup$ It's not MERELY "convenient". As the example in my answer shows, it is sometimes necessary for the correctness of standard formulas. $\endgroup$ – Michael Hardy Jan 17 '17 at 22:43
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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.

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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 $m \in \mathbb{N}$.

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  • $\begingroup$ Made an attempt to answer with consideration to the comments from @FedericoPoloni, let me know if you have any comments! $\endgroup$ – Toivo Säwén Jan 19 '17 at 9:15

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