According to my knowledge factorial means that the number is multiplied by the numbers preceding it up to $1$ . That's it . $5!=5×4×3×2×1=120$



$n!=n×(n-1)×\cdots \times 2×1$

And similarly


But this is not possible as we have crossed 1 before and another 1 is not possible . So it should be ND. Please tell me where I am wrong in this?

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    $\begingroup$ At an elemntary stage,0! Is only defined to be that way since it is convenient in the formulas and follows the pattern of ${n!\over {n}} = (n-1)!$, but after that ,the generalization of this,the gamma functuon is usually used. $\endgroup$ Jan 4, 2017 at 12:33
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    $\begingroup$ @LoganLuther At an elementary stage, $0!$ makes perfect sense both as the number of permutations of a set of no objects, and as the empty product $\prod_{i = 1}^0 i$. $\endgroup$
    – Arthur
    Jan 4, 2017 at 12:38
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    $\begingroup$ @Arthur the result is usually controversial,as to what defintion of $n!$ we are using,by the OP's defintion,there is no mention of the combinatoric meaning of $n!$,and I based my comment on that. $\endgroup$ Jan 4, 2017 at 12:43
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    $\begingroup$ @LoganLuther Then at the very least, you have the empty product. There is no controversy about that, except from people who have just learned about it. $\endgroup$
    – Arthur
    Jan 4, 2017 at 12:44
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    $\begingroup$ Possible duplicate of What is $n!$ when $n=0$?. In fact, this is a duplicate of a duplicate. $\endgroup$ Jan 4, 2017 at 13:51

5 Answers 5


"According to my knowledge factorial means that the number is multiplied by the numbers preceding it up to 11 . That's it . 5!=5×4×3×2×1=120"

If you are certain that this is the definition of factorial, then you're certainly correct that $0!$ should be undefined.

But here's something closer to the truth: mathematicians observed that products of sequences of numbers kept arising in things they were working on. In particular, when you asked "How many ways are there to order 3 items?' you ended up with

1 2 3
1 3 2
2 1 3
2 3 1
3 1 2
3 2 1

and there turned out to be $6 = 3 \times 2 \times 1$ such ways. And in general, when you considered all possible permutations of $n$ items, there turned out to be $n \times (n-1) \times \ldots \times 2 \times 1$ of them. It seemed as if having a name for this oft-occurring product would be a Good Thing, and someone said "let's write $n!$ for that." (The actual truth is surely more complicated that this, but I'm trying to give you the gist). In particular, new notations and definitions come up via utility; they're not handed down on stone tablets from above.

Then, perhaps, someone noticed that there are several ways to take $k$ things from a group of $n$, and that the number of such ways turned out to be $$ \frac{n!}{k! (n-k)!} $$ but that this formula unfortunately only worked in the case where $0 < k < n$, but not for $k = 0$ or $k = n$. But since it made perfect sense to take $n$ things from a pile of $n$ things (and indeed, there was only one such way), that person realized that if the definition of $n!$ was extended to include $0! = 1$, the formula would still work. (And through good fortune, it also worked for the $k = 0$ case as well with this new definition.)

At that point, a bunch of mathematicians probably started arguing, and after a while, the $0! = 1$ crowd won out, and we all agreed that henceforth we'd use that definition. It was nice that $0!$ turns out to be the number of ways of arranging 0 items as well (since there are no choices to make, there's only one possible way).

So: your original model of the definition of $n!$ is perhaps flawed, but if you want to believe it, you can, and can say "I personally will never write $0!$, because it's not defined." All your formulas will have special cases when $0$ comes up, but that's OK. The rest of us will use the other definition and have simpler formulas. When you look at our formulas, you can scoff, but you'll also know what they mean, because we've told you that for us, $0!$ is just another notation for the number $1$.


I am not aware if you have had any exposure to permutations and the use of factorials,but I first give an incomplete answer before it,and then elaborate with the combinatorics.

First Answer:If we are to define $n!$ in the way you described,we can notice this Recursive formula:$${n! \over n}=(n-1)!$$

For example:$${3! \over 3} =2=2!$$

This looks like a good property to make our defintion more mathematical. Thus We can use it as The $definition$ of $n!$.

Now our first definition excluded $0!$,but can we find it's value now?

If we put $n-1 =0$ in our New defintion,we get:$${1! \over 1}=1=0!$$

So our formula suggests that this quantity is indeed 1.

BUT,There is an issue,we could,instead of substituting $n=1$, substitute $n=0$, to find $0!$, but this comes out as undefined:$${0! \over 0}=-1$$, which can have no value,therefore this argument while interesting,is not flawless.

Second Answer:Suppose you have 3 coins,and you want to arrange them in a line, in how many ways you can do that? Well you could just count them,but what about 20 coins? The answers will get gargantuanly big(as you will see),so let us be smart and find another way to count them.

Now ,suppose we marked each position with a letter,$a,b,c,... \mathrm {etc}$.and we have $n$ number of coins.

How many different coins can we put on position $a$?

Well we have $n$ number of distinct coins, so the number will obviously be n.

What about the position $b$?

We have already used 1 coin(we just do not know which), so the answer is $n-1$.

Now,how will we be able to combine these results to get an answer?

Well,think of it as a tree and it's branches. We first have a point,for position $a$ and out of which comes out $n$ different branches.

What about the position $b$? We have after each branch the point for position $b$ , meaning we want to determine how many different branches will come out i.e how many different number of coins can we put in there? Well, we already know the answer is $n-1$, so there will be $n-1$ new branches after the previous $n$ branches that we had,so there are $n$ "$n-1$ " branches,or :$$n×(n-1)$$

We continue like this until we get to the last position ,and there will be no positions to evaluate further.

At each stage,the next set of branches had one less branch than the previous set,and they were being multiplied together, so the final answer will be:$$n×(n-1)×...× 2×1$$ This is just our formula for $n!$ !!! So we have a new tool to define it.

$$\mathrm {Defintion}:$$,$n!$ is the number of ways that $n$ distinct objects can be arranged in a line.

Now let us tackle $0!$ in this way,

How many ways are there to arrange $0$ distinct objects?

The answer is well ,$1$. Here it is .It has already been done. I have had No objects and I have arranged them in that way!!.

(I understand that it might be at first not obvious as to why the above statement is true,but there is no other answer besides $1$,but, I will leave that to you to realize.)

ADD:As I and others have mentioned,the easiest most convenient way that has no hazzy arguments,is to just define it to be $1$...

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    $\begingroup$ I know both of these explaination. I have done a small research before asking it in SE . ... But can you tell me where my method is not correct. $\endgroup$
    – Prabhat
    Jan 4, 2017 at 13:42
  • $\begingroup$ You do not define any way to calculate $0!$ ,and you are correct that with your defintion,$0!$ will be indeed undefined,and in my answer I try to extend the definition in such a way as to allow $0!$to be defined. $\endgroup$ Jan 4, 2017 at 13:45

You have identified the problem with zero. Our usual definitions of mathematical operations are often difficult to understand when we try to apply them to 0. For example, while $x^4 = x\cdot x\cdot x\cdot x$ or more concretely $2^4 = 2\cdot 2\cdot 2\cdot 2$, we are stuck defining $0^0 = 1$. Likewise, we can say $10/2 = 5$ since $2\cdot 5 = 10$, but we are stuck defining $10/0$ which can't be defined.

To handle $0!$, we are similarly stuck defining $0! = 1$. This is not because of a deep mathematical reason, but instead is a necessary extension of our definition of factorial.

  • $\begingroup$ Ha my bad. Spent too much time on physics forums tonight. $\endgroup$ Jan 4, 2017 at 12:41
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    $\begingroup$ 0^0=not defined I think $\endgroup$
    – Prabhat
    Jan 4, 2017 at 13:20

Many formulas are simpler if we define 0! to be 1. If we defined 0! to be 0 in many cases we have countless solutions.

Let one simple example according to you,

$\binom{6}{6} = \frac{6!}{6! \times 0!}$

= $ \frac{1}{0 \times -1 \times -2 \times -3 ...... }$

= $ \frac{1}{0} = \infty$

And in many cases we get always 0 as the answer.

Example in how ways the word STACK arranged.

We have $5! = 5 \times 4 \times 3 \times 2 \times 1 \times 0 \times -1 \times ....... = 0$

Hope its help you.

  • $\begingroup$ ConsiderInG these answers I think that we have to take 1 as 0! . No other value will satisfy our problems .. then this method should apply to every subject but I didn't see any one alike to this case. $\endgroup$
    – Prabhat
    Jan 4, 2017 at 13:24
  • $\begingroup$ Then tell me other cases where this method is not apply. $\endgroup$ Jan 4, 2017 at 13:32
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    $\begingroup$ Like 0^0 I think that it is equal to 0 as 0^0=0/0=0×ND=0. BUT IT IS NOT TRUE . 0^0 is indeterminate $\endgroup$
    – Prabhat
    Jan 4, 2017 at 13:36
  • $\begingroup$ Yes I agree with these cases. We have different results in different cases. But in factorial we have as I explain above. $\endgroup$ Jan 4, 2017 at 13:45

My favorite is the following. We define $n!$ to be the number of ways $n$ distinct objects can be ordered. Now, how many ways can you order $0$ objects? Well, you can either say that (1)"you can't", (2) "there are zero ways", or (3) "there is one way to organize zero objects". If you choose (1) you're just being defiant. Moreover, you would like to avoid zeroes as much as possible, since factorials do indeed appear in denominators so you try to find a way around (2). So we go with definition (3), because we choose to come to the agreement that there is one way to order zero objects.


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