I don't know what operation to preform with "!"

Question

I am currently working on my math homework (the book was optional), and I don't know what to do when I see the "!". I am trying to evaluate a problem with this symbol.

Answer 1

Assuming it isn't an exclamation point, $!$ is a factorial. That is for a positive integer $n$

$$n!=n(n-1)(n-2)\cdots 2 \cdot 1.$$

For instance $5!=5\cdot4\cdot3\cdot2\cdot1=120$.

Answer 2

You may want to prove that the number of different subsets with $\,k\,$ elements that a set with $\,n\,$ elements , $\,0\leq k\leq n\,$ , is given by what's called **the binomial coefficient$$

$$\binom{n}{k}:\stackrel{\text{Def.}}=\frac{n!}{k!(n-k)!}$$

If the above is way too hard, you can try to work out different cases: for any $\,n\,$, the cases $\,k=0,1,2,,n-2,n-1,n\,$ . On purpose I recommend you these ones as you will find something really nice, and sometimes surprising at the beginning, about these cases. Perhaps first choosing, say $\,n=3,4\,$ , then something more general.

Answer 3

There are several rather common operations associated with $!$. By the far the most common is the factorial which is defined for natural numbers as $$n! = n\times (n-1)!\ \ \ \text{with}\ \ \ 0! = 1$$ This is simply a product of all the natural numbers from $1$ to $n$. Incidentally, this is also the number of ways to rearrange $n$ distinct objects into order.

For your interest, another commonly used and related notation is the double factorial $n!!$ This is often used to denote the product of the odd numbers from $1$ to $n$.

Yet another one which uses an exclamation mark (although this one is much rarer from my experience) that I have seen is the number of derangements or the subfactorial. This is sometimes denoted $!n$ and is given as $$!n = n\times !(n-1) + (-1)^n\ \ \text{with}\ \ \ !0 = 1$$