Simplification with factorials

Question

If I have the fraction $\dfrac{(2n)!}{2!((2n)-2)!}$, am I allowed to factor out the $2$ so that the equation becomes $ 2 \cdot \dfrac{n!}{1!(n-1)!}$? If not, is there anything I can do with it?

Edit: clarified, changed $2n!$ to $(2n)!$

Answer 1

Note that

$$\frac{(2n)!}{2!(2n-2)!}=\frac{(2n)(2n-1)(2n-2)!}{2!(2n-2)!}=n(2n-1)$$

Answer 2

$$(2n)!=2n\cdot(2n-1)\cdot\ldots\cdot(n+1)\cdot n!$$

Answer 3

No, that doesn't work. You are making numerous mistakes here! It seems that you are doing something like:

$$\frac{2n!}{2!(2n-2)!} \overset{\text{'take out 2's'}}{=} 2\cdot \frac{\not{2}\cdot n!}{(\not{2}\cdot 1!)(\not{2}\cdot (n-1)!)}= 2\cdot \frac{n!}{1!(n-1)!}$$

OK, I can see three mistakes:

First of all, if both the numerator and denominator are divisible by $2$, then you remove the $2$ on both sides, but you do not add a $2$ to the front; that's not what we mean by 'take out a $2$'

Second, you are trying to get a $2$ out of every term, i.e. take a $2$ out of $2n!$, out of $2!$, and out of $(2n-2)!$. But, that means that you get two $2$'s in the denominator, rather than one. And you can only get rid of one given that you would have only one $2$ in the numerator. That is, it is not true that:

$$\frac{2a}{2b\cdot2c}=\frac{a}{bc}$$

but rather you get:

$$\frac{2a}{2b\cdot2c}=\frac{\not{2}a}{\not{2}b\cdot 2c}=\frac{a}{2bc}$$

Third, you are assuming that $(2n)!=2 \cdot n!$, which is not true. You have that:

$$(2n)! = (2(n))!$$

but that is not the same as $2\cdot n!$. You're effectively assuming that $(a \cdot b)! = a \cdot b!$, which you can easily verify is not the case.

Likewise, you cannot say that $(2n-2)! = 2\cdot (n-1)!$