**Variant:**
In a class with $2n+1$ students, each student is allowed to choose a number from the following list: $n, n+1, -n, -n-1$. The teacher then collects all chosen numbers and calculates the sum. How many ways are there for the sum to be zero?

**Solution**
Suppose the number of students who chose 100,101,102,300,301,302 are $a,b,c,d,e,f$ respectively, so we have:
$$a+b+c+d+e+f = 10$$
and
$$100.a + 101.b + 102.c + 300.d + 301.e+ 302.f = 2012$$

Note that we can write 100=201-100-1, 101 as 201-100, 102 as 201-100+1, 300 as 201+100-1, 301 as 201+100, 302 as 201+100+1, so the second equation becomes: $$201(a+b+c+d+e+f) + 100(d+e+f-a-b-c) + (c+f-a-d) = 2012$$ Since $a+b+c+d+e+f = 10$ then: $$100(d+e+f-a-b-c) + (c+f-a-d) = 2$$

Now, since $|c+f-a-d| \leq 10$ then it must be the case that: $$d+e+f-a-b-c = 0$$ and $$c+f-a-d = 2$$

We also note that when the students choose one of the numbers, they're essentially making two choices: first, they choose if they want to choose small or big numbers (be included in $a+b+c$ or $d+e+f$), and second, they choose if they want their number to end with 0,1, or 2 (be included in $a+d, b+e$ or $c+f$).

Thus, the number of ways to satisfy the first constraint $a+b+c = d+e+f = 5$ is 10C5, which means 5 students choose the small numbers and the rest choose the big numbers.

In order to satisfy $c+f - a - d = 2$, we make the following substitutions: $x = a+d, y = b+e, z = c+f$. Then $x+y+z = 20, z-x = 2$. We divide the cases by the values of $z$. Note that once a student chooses to be included in $x$ or $y$ or $z$, he only needs to choose if he wants to choose a big or small number, since each of $x,y,z$ contains exactly one big and one small number.

If $z > 6$ then $x > 4$, impossible.

If $z = 6$, then $x = 4, y = 0$. There are 10C6 = 210 ways to choose this.

If $z = 5$, then $x = 3, y = 2$, there are 10! / 2! 5! 3! = 2520 ways.

If $z = 4$, then $x = 2, y = 4$, there are 10! / 4! 4! 2! = 3150 ways.

If $z = 3$, then $x = 1, y = 6$, there are 10! / 3! 6! 1! = 840 ways.

If $z = 2$, then $x = 0, y = 8$, there are 10C8 = 45 ways.

If $z < 2$ then $x < 0$, impossible.

So the total number of ways is $$_{10}C_{5} (210+2520+3150+840+45) = _{10}C_{5} 6765 = 1 704 780$$

**Solution to variant**

Suppose the number of students who chose $-n-1, n+1, -n, n$ are $a,b,c,d$ respectively, so we have: $$a+b+c+d = 2n+1$$ $$(n+1)(b-a) + n(d-c) = 0$$

Note that because $n$ and $n+1$ are relatively prime, so we must have $n+1 | d-c$. But $|d-c|$ can take any values from $0$ to $2n+1$, so $d-c = 0, n+1, -(n+1)$.

If $d-c=0$, then $b-a=0$, which means $a+b+c+d$ would be even, impossible.

If $d-c=n+1$, then $b-a = -n$. Combined with $a+b+c+d = 2n+1$ we have $(a,b,c,d) = (n,0,0,n+1)$. (Remember that $a,b,c,d$ have to be non-negative).

This means that $n$ of the students choose $-n-1$ and the other $n+1$ choose $n$. There are $_{2n+1}C_{n}$ ways to do that.

The case of $d-c = -n-1$ is similar, where $(a,b,c,d) = (0,n,n+1,0)$, and there are $_{2n+1}C_{n}$ ways. Note that all these ways are disjoint, so the total number of ways is $ 2_{2n+1}C_{n}$

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