Fine; MrGenius you've got

$$\textcolor[rgb]{1,0,0}{20/20}$$

CONGRATULATIONS !    

Here's your cookie:

It's because $$(n-1)!=\frac{n!}{n}$$

That means :

$$\\6!=1\times 2\times 3\times 4\times 5\times 6=720 \\(6-1)!=5!=\frac{6!}{6}=120 \\(5-1)!=4!=\frac{5!}{5}=24 \\(4-1)!=3!=\frac{4!}{4}=6 \\(3-1)!=2!=\frac{3!}{3}=2 \\(2-1)!=1!=\frac{2!}{2}=1 \\\textcolor[rgb]{0,0,1}{(1-1)!=0!=\frac{1!}{1}=1}$$

You may be right; wait, I'll remove things in the wording...

The 1st sum is 1-1+1-1+1-1+... and not 1+0+1+0+1+0+...

I didn't; I didn't gave the mass of 2L of water, nor the molar mass of water; I have just written down the formulas used to calculate this.

In order to calculate the amount of matter, you first have to determinate the mass m of 2L of water, using the first formula given, then you have to determinate the molar mass M of the water, and finally use the second formula to find the value of n.

You might think that all that calculation are wrong; but they AREN'T.

0! equals 1;

One of the ways to show this is to "complete the pattern":

$$\\5!=1\times 2\times 3\times 4\times 5=120 \\4!=\frac{5!}{5}=\frac{120}{5}=24 \\3!=\frac{4!}{4}=\frac{24}{4}=6 \\2!=\frac{3!}{3}=\frac{6}{3}=2 \\1!=\frac{2!}{2}=\frac{2}{2}=1 \\0!=\frac{1!}{1}=\frac{1}{1}=1$$

But that's where the pattern ends; you can't continue, because you end up with this:

$$(-1)!=\frac{0!}{0}=\textcolor[rgb]{1,0,0}{\frac{1}{0}}$$

See what I mean ?

Precisely.

The answer is $${\mathtt{\,-\,}}{\frac{{\mathtt{1}}}{{\mathtt{12}}}}$$ ; it can seem strange, because it's a negative number, and it's not an integer; but that result is absolutely true.

Here's your cookie:

There's the proof I usually use:

$$\\S_1=1-1+1-1+1-1+... \\S_2=1-2+3-4+5-6+... \\S=1+2+3+4+5+6+... \\\\S_1=1-1+1-1+1-1+... \\1-S_1=1-(1-1+1-1+1-...)=1-1+1-1+1-1+...=S_1 \\1=2S_1 \\S_1=\frac{1}{2}$$

$$\\\\S_2=1-2+3-4+5-6+... \\S_2+S_1=(1-2+3-4+5-6+...)+(1-1+1-1+1-1+...) \\S_2+S_1=2-3+4-5+6-7+... \\-1+S_1+S_2=-1+(2-3+4-5+6-7+...)=-1+2-3+4-5+6-7+...=-S_2 \\-1+\frac{1}{2}=-2S_2 \\-2S_2=-\frac{1}{2} \\S_2=\frac{-\frac{1}{2}}{-2}=\frac{1}{4}$$

$$\\\\S=1+2+3+4+5+6+... \\S-S_2=(1+2+3+4+5+6+...)-(1-2+3-4+5-6+...)=4+8+12+16+20+24... \\=4(1+2+3+4+5+6+...)=4S \\-S_2=3S \\-\frac{1}{4}=3S \\S=-\frac{-\frac{1}{4}}{3}=-\frac{1}{12}$$