Which rule is applied to define the operator precedence for factorial

Question

Please apologize the question, I struggled with finding a good formulation in the first place:

Looking at $\binom{2n}{k}$ it is very clear that for n,k integer and n>k we can solve it by calculating:

$$(2n)!\over k!(2n-k)!$$

$2n!$ is clearly not the same as $(2n)!$ as for $n=3$ we easily see $2\times3\times2\times1 \neq 6\times5\times4\times3\times2\times1$

Which mathematical rule is responsible for this? I mean it is clearly just a substitution, but still I struggle in finding any ruling for this differentiation - probably because it is as basic.

Answer 1

The standard notation for the factorial function is rather unusual.

Usually we define a function $f$ from space $A$ to space $B$ by saying $f: A\rightarrow B$ and then whenever we write the function we (usually) require parentheses to describe precisely what the function operates on, so $2f(a)$ and $f(2a)$ mean clearly different things. In the order of operations you apply parentheses first and then apply all functions (the introductory form of the order of operations: PEMDAS can be broadened to include all functions in the E for exponentiation - all functions would take the same precedence as exponentiation in an analogous way to how the M for multiplication and D for division take the same precedence as each other).

The "$!$" notation is usually taken as a function whose argument is the variable or constant which is written directly in front of it.

You might be interested in further reading: Most ambiguous and inconsistent phrases and notations in maths.

Answer 2

The factorial function $x \mapsto x!$ is by convention written in "postfix" notation, like exponentiation $x \mapsto x^n$. The convention is that these postfix operators have high precedence (i.e., you evaluate them before other operators unless parentheses tell you otherwise): $3x^2$ means $3 (x^2)$ not $(3x)^2$ and $2n!$ means $2(n!)$ not $(2n)!$.