How to interpret $(2n)!$

Question

It's all in question: how to interpret the factorial from $2n$? Is $(2n)!$ equal to $n!\times n!$ ?

The problem is in Combinations if the combinations is $\binom{2n}3$.

P.S.

The main problem is how to write $C^3_{2n}$ (C is for combinations in combinatiorics)

Answer 1

The rules of writing mathematics state that if you do not enclose the factor 2 in parentheses, then $2n!$ is interpreted as $$2n!=2\times(n!)=2\times1\times2\times3\times\cdots\times n.$$ The factorial of $2n$ has to be written as $(2n)!$. Using the definition of the factorial, we have

$$(2n)!=1\times2\times3\times\cdots\times(2n-1)\times(2n).$$

So to answer your question, the factorial of $2n$ that enters in the expression $\binom{2n}3$ is not equal to $n!\times n!$ (which is written $n!^2$)... it is a larger number (can you see why ?).

Note: The combination (or binomial coefficient) $\binom{2n}{3}$ is equal to $$ \binom{2n}3=\frac{(2n)\times(2n-1)\times(2n-2)}{1\times2\times3}.$$

Answer 2

$(2n)! = 1\cdot2\cdot3\cdot\cdots\cdot (2n-1)\cdot 2n$, as usual.

Answer 3

Interpret it as

$$1\cdot2\cdot\ldots\cdot(n-1)\cdot n\cdot(n+1)\cdot\ldots(2n-1)\cdot2n.$$

Verify that this is different from $n!\cdot n!=(n!)^2$ instead.