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## Main Question or Discussion Point

Why is zero factorial equal to 1?

- Thread starter quasi426
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Why is zero factorial equal to 1?

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As they say **:**

!

*__Edit__: Sorry about that! Here's why:

Three reasons, from my perspective:

1) 0! is**defined** to be equal to one.

2) (This will sound weird. Oh well, here goes) You know how

[tex]\frac{{n!}}{n} = \left( {n - 1} \right)! [/tex]

---->

And so,

[tex]n = 1 \Rightarrow \frac{{1!}}{1} = 0! = 1 [/tex]

3) Consider combinatorials, for example. How many ways are there to choose [tex]k[/tex] objects out of a total quantity of [itex]n[/itex] objects?

*We would use the expression:

[tex] \frac{{n!}}{{k!\left( {n - k} \right)!}} [/tex]

So, for example, there are 6 ways to choose 2 objects out of a set of 4.

But what if [tex]k = 0[/tex], --->i.e., we choose__nothing__?

[tex]\frac{{n!}}{{0!\left( {n - 0} \right)!}} = \frac{{n!}}{{0!n!}} = \frac{1}{{0!}} = 1\;({\text{or else we have a problem}}) [/tex]

You see, there only**one way** to choose **nothing**.

-----------------------------------------------------------

Personally,

I usually just go with the first statement:

-"0! is**defined **to be equal to one"-

___________________________"Man has pondered

Since time immemorial

Why 1 is the value

Of zero-factorial."

!

*

Three reasons, from my perspective:

1) 0! is

2) (This will sound weird. Oh well, here goes) You know how

[tex]\frac{{n!}}{n} = \left( {n - 1} \right)! [/tex]

---->

And so,

[tex]n = 1 \Rightarrow \frac{{1!}}{1} = 0! = 1 [/tex]

3) Consider combinatorials, for example. How many ways are there to choose [tex]k[/tex] objects out of a total quantity of [itex]n[/itex] objects?

*We would use the expression:

[tex] \frac{{n!}}{{k!\left( {n - k} \right)!}} [/tex]

So, for example, there are 6 ways to choose 2 objects out of a set of 4.

But what if [tex]k = 0[/tex], --->i.e., we choose

[tex]\frac{{n!}}{{0!\left( {n - 0} \right)!}} = \frac{{n!}}{{0!n!}} = \frac{1}{{0!}} = 1\;({\text{or else we have a problem}}) [/tex]

You see, there only

-----------------------------------------------------------

Personally,

I usually just go with the first statement:

-"0! is

Last edited:

-Job-

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Of course consider, for n distinct objects, there are n! permutations of those objects. So it seems that for 0 objects there's 1 possible permutation.

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Right, as according

http://mathworld.wolfram.com/Zero.html

http://mathworld.wolfram.com/Zero.html

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Is this the correct definition of factorial, or is it inconsistant with 0! ?

[tex]x! = \prod_{n=1}^{x} n[/tex]

[tex]x! = \prod_{n=1}^{x} n[/tex]

Last edited:

shmoe

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This is the* definition when x is a positive integer. With the usual convention that an empty product is 1, it is consistant with our usual definition for 0! when x=0. However, x=0 is usually left as a special case and explicitly defined as 0!=1.Joffe said:Is this the correct definition of factorial, or is it inconsistant with 0! ?

[tex]x! = \prod_{n=1}^{x} n[/tex]

*there's more than one equivalent way to define factorial of course

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For instance, if 0! was anything other than 1, the cosine function wouldn't make any sense. Consider: f(x) = Cos(x) = x^0/0! + x^2/2! -x^4/4!... where x is a radian measure. So, if 0! was not equal to 1, then the first term in the series would not equal 1, and the Taylor series that derived it would be wrong, which would turn everything that we know about math upside down.

shmoe

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It would just change how we express the power series, it would not have any effect whatsoever on the mathematics. This is the same situation as adopting the convention x^0=1 for x=0, without this your power series as expressed doesn't make sense either. This is not a mathematically compelling reason for either convention/definition, rather a notational one.Quadratic said:For instance, if 0! was anything other than 1, the cosine function wouldn't make any sense. Consider: f(x) = Cos(x) = x^0/0! + x^2/2! -x^4/4!... where x is a radian measure. So, if 0! was not equal to 1, then the first term in the series would not equal 1, and the Taylor series that derived it would be wrong, which would turn everything that we know about math upside down.

jcsd

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This holds true even when |A|=0 as B in this case contains the empty function (and the empty function only), so you can see there is justification for definition 0! = 1.

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[tex]\Gamma{\left(1\right)}=0!=\int_0^{\infty}e^{-x}\,dx=1[/tex]

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I know that the gamma function can be used to solve factorials for any number but are there any other special rules like this one?

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from the MathWorld

matt grime

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Step back, take a breath and try to think what the person who first came up with such Taylor/McLaurin series might have done if 0! weren't 1, because surely he or she wouldn't have used something that was undefined or unsuitable?Quadratic said:For instance, if 0! was anything other than 1, the cosine function wouldn't make any sense. Consider: f(x) = Cos(x) = x^0/0! + x^2/2! -x^4/4!... where x is a radian measure.

benorin

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25 < my age < 54. Start from the number of my age and toss a fair coin, plus one if it is a head and minus one otherwise, on average, you need to toss 210 times to hit the bound, i.e. 25 or 54. There are two integers satisfying the condition. Fortunately, the smaller one is my age.

HallsofIvy

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And I'm sure the gamma function is fascinated by you.

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could you show me some basic properties of factorials? is factorial distributive over product?

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Can't you test that hypothesis yourself?

What are 6!, 3!, and 2! equal to?

What would 3!*2! be?

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