 Could someone explain the the ! symbol in mathematics…what it means and how it is used?

I had always thought it stood for the number of combinations that could be formed from a given number of objects (e.g. the number of different “words” that could be formed from a 26-letter alphabet) although I can’t remember where or when I heard that. Recently an acquaintance told me that I was misinformed and tried to explain what it really meant. Unfortunately he has an advanced degree in physics; I do not. I nodded and smiled and tried to look enlightened, but actually he lost me at the first curve.

Can someone explain it in (relatively) easy-to-understand terms?

It means factorial and is not that difficult to understand.

2! = 12 = 2
3! = 1
23 = 6
4! = 1
234 = 24
5! = 12345 = 120
6! = 12345*6 = 720

And so on.

It’s because we’re constantly amazed at numbers (!!)

Actually, it’s the factorial sign. It’s easiest to give an example 5!, pronounced “five factorial”, is equal to 5 X 4 X 3 X 2 X 1. This sort of thing comes up pretty often, especially in nstatistics, combinatorials, and infinite series.
A double exclamation point means you skip every other number:

5!! = 5 X 3 X 1

You can even have something like this for non-integers, The factorial of x is equal to the Gamma Function of x + 1, and the Gamma Function is defined for non-integer values

It means Factorial, the product of all all positive integers less than or equal to the number.

E.g. 4! = 1 x 2 x 3 x 4

One of its uses are calculating the Binomial coefficient, which sounds like what you were thinking of.

C(n, k) = n! / ( k! x (n-k)! )

Eg. how many poker-hands exists?

We draw 5 cards from a deck of 52.

C(52, 5) = 52! / ( 5! x (52-5)! ) = 2,598,960

Lots more in wikipedia!

It’s six of one thing, 3! of the other.

Yes, factorial. You may have been confused because it is often used in calculating combinations.

You’re almost right. The actual definition has already been given, which is about multiplying together the number and all the other numbers between it and one.

But it turns out that if you want to arrange n different things in order, there are n! ways to do it. So, if you have 26 (different) cards, there are 26! ways they could be stacked (or, to put it another way, there are 26! possible results of shuffling a 26-card deck). That’s not the definition, but it is true.

If you’re making words from an alphabet, that’s different, though, because you can use the same letter twice. (If you have two cards, “A” and “B”, you can shuffle them two ways: “AB” or “BA”, but if you’re making two-letter words using ‘a’ and ‘b’, you can make ‘ab’, ‘ba’, ‘aa’ or ‘bb’)

But 26! actually is the number of 26 letter “words” you could make that used each letter exactly once.

0! = 1
n! = n x (n-1)!

(undefined for n < 0)

As the previous posters have indicated, N! is the number of permutations or ordering for N objects, so 3!=123=6 because there are 6 orderings of (1,2,3)

123
132
213
231
312
321

So by and large you were correct.

Now reading between the lines of your conversation and realizing that advanced math and physics majors like to do (myself included) like to show off, I suspect what your friend was trying to tell you about was the Gamma function referenced by CalMeacham.

This is a function that is defined in terms of an integral and has the property that for any non-negative N!=Gamma(N+1). So Gamma(5)=4!=123*4. But the Gamma function is a continuous function so that you can define Gamma(4.5)=3.5! which unfortunately can’t be expressed in terms of a simple multiplication but which happens to be 11.63 (somewhere between 3!=6 and 4!=24.

Interestingly (ie I want to show off now) the Gamma function is also defined for most numbers less than 1 (the equivalent of negative factorials). But is not defined at 0 or at the negative integers.

Correct. So there are 52! possible results in shuffling a full deck of cards. 52! is greater than a one followed by 67 zeroes.

Remember that the next time you read creationist arguments about how something more unlikely than, say, one in a trillion trillion can never happen. Something WAY more unlikely than that happens every time you shuffle a deck of cards.

If you shuffle a deck thoroughly, then you have produced something that the world has never seen before, and if left only to chance, will never see again. If a trillion people shuffled a trillion decks of cards each, a trillion times per second, and had been doing so since the universe began 13 billion years ago, the chance that one of them would have produced the same order of cards you did would be about one in a trillion trillion.

As the previous posters have indicated, N! is the number of permutations or ordering for N objects, so 3!=123=6 because there are 6 orderings of (1,2,3)

123
132
213
231
312
321

So by and large you were correct.

Now reading between the lines of your conversation and realizing that advanced math and physics majors like to do (myself included) like to show off, I suspect what your friend was trying to tell you about was the Gamma function referenced by CalMeacham.

This is a function that is defined in terms of an integral and has the property that for any non-negative N!=Gamma(N+1). So Gamma(5)=4!=123*4. But the Gamma function is a continuous function so that you can define Gamma(4.5)=3.5! which unfortunately can’t be expressed in terms of a simple multiplication but which happens to be 11.63 (somewhere between 3!=6 and 4!=24.) The main property of the Gamma function is that Gamma(N)=(N-1)*Gamma(N-1)

Interestingly (ie I want to show off now) the Gamma function is also defined for most numbers less than 1 (the equivalent of negative factorials). But is not defined at 0 or at the negative integers.

Sorry about the double post (darn hamsters).

And, and, is defined for all complex numbers, except non-positive integers.

I’d say the OP was probably right, but mis-remembered. N! is the number of words of length N made up of the same N distinct letters. That is, 4! is the number of different ways to make a four-letter word from ABCD, even though the words are nonsense, and don’t include examples like AAAA.

Your guess wasn’t that much off the wall, though, since factorial expressions can (must?) be used in calculating the number of possible choices or subsets of size r which you can choose out of n objects. There’s a different symbol for that, read as “n choose r”, but I don’t know how to type it.

[sup]n[/sup]C[sub]r[/sub].

Written as [noparse][sup]n[/sup]C[sub]r[/sub][/noparse].

I know it is fairly straightforward, but stuff like this just never ceases to amaze me. I had never thought about that.

Kind of like when you take a piece of paper and fold it in half 100 times (doesn’t seem all that much) you end up with a stack of paper higher than the outer reaches of the universe, more than 15 billion light years.

A) Help correct the inertia of history… write n! (or, if you prefer prefix notation, Fact(n) or such things) whenever natural, instead of Gamma(n+1). Gamma(n+1) is a stupid conflicting convention which we have no need to keep burdening ourselves with.

B) People are often introduced to the generalized (i.e., non-integer) factorial in terms of an integral which happens to compute it, but I don’t believe this is a very good way of motivating it as the natural extension of the factorial function.

Instead, I would motivate it this way: note that, even knowing nothing about what the factorial function is, we have that x! = x!/(n+x)! * (n+x)!/n! * n!, for every n.

With the defining properties that 0! = 1 and y! = (y - 1)! * y, we automatically know that n! = 1 * 2 * … * n and that (n + x)!/x! = (1 + x) * (2 + x) * … * (n + x), whenever n is a natural number. So our equation becomes x! = [1/(1 + x) * 2/(2 + x) * … * n/(n + x)] * (n + x)!/n!, for every natural number n. The only part we don’t know how to calculate right off-the-bat for non-integer x is that last (n+x)!/n!.

But: for the standard integer factorial, we know that (n + x)!/n! is approximately equal to n^x, whenever n is much larger than x. (We can make this precise in terms of limits (the ratio between these goes to 1 as n goes to infinity), and can even derive it solely from the “log-convexity” of the factorial function). If we choose to preserve this property even for the generalized factorial, we’re done:

We will have that x! must equal 1/(1 + x) * 2/(2 + x) * … * n/(n + x) * n^x in the limit as n goes to infinity; this defines the factorial even for non-integer x. Furthermore, our argument shows that this is the unique function with the desired properties (that 0! = 1, that n! = n * (n - 1)!, and that (n + x)!/n! ~ n^x for large n (which can be derived from log-convexity on positive inputs)). Thus, any function with the same properties is equal to this one.

In particular, the usual integral definition everyone sees can be shown to also have the same properties, and therefore is automatically equal to this one, but I think it’s much more intuitive and instructive to see where this one comes from. (Plus, the uniqueness element of the argument provides a nice proof that (1/2)! = 1/2 * sqrt(pi) as well, but I’ll save that for later…)

brocks writes:

> If you shuffle a deck thoroughly, then you have produced something that the
> world has never seen before, and if left only to chance, will never see again. If
> a trillion people shuffled a trillion decks of cards each, a trillion times per
> second, and had been doing so since the universe began 13 billion years ago,
> the chance that one of them would have produced the same order of cards you
> did would be about one in a trillion trillion.

Um, no. Let me explain:

First, 52! is a little less than (10^68), so there are that many different orderings of the cards.

A trillion is 10^12. If a trillion people shuffled a trillion decks of cards each a trillion times a second, that would mean that there would be

(10^12) X (10^12) X (10^12) = (10^36)

different orderings of the cards per second.

The number of seconds in 13 billion years is

1.3 X (10^10) X 3.65 X (10^2) X 2.4 X (10^1) X 3.6 X (10^3)

so there are a little more than (10^17) seconds in 13 billion years.

This means that the total number of different orderings these people would have produced is (10^53).

You seem to think that it would be necessary to produce about (10^68) different orderings to expect to see even one case where two of those orderings match. That’s not true. It’s only necessary to produce about (10^34) different ordering to expect to see about one case where two of those orderings match.

This is an example of what’s called the Birthday problem:

http://en.wikipedia.org/wiki/Birthday_problem

Suppose you want to pick a group of people just barely large enough that you can expect one pair of those people to have the same birthday. You don’t need 365 people. You only need 23 people. Approximately speaking, if there are n different values that something can take, you only need a group of about the square root of n different things chosen randomly for there to be about one match among them. (Well, it’s not exactly the square root, but since we’re only dealing with numbers to the nearest power of 10, that’s close enough.) So in the situation you describe, there would be an enormous number of matching pairs of orderings.

Someone please check my arithmetic, since it’s easy to mess up in cases like this.

I don’t know if anyone’s mentioned it, but ! represents the factorial function.