We want to pick out the same number of ~0~ as ~1~. Let ~a~ be the number of ~0~ and ~b~ be the number of ~1~. Let ~m~ be the larger of ~a~ and ~b~ and ~n~ be the smaller of the them.

Since we want the same number of ~0~ and ~1~, we should pick out the same number from of the total number ~0~ and ~1~. For one taking out ~i~ ~0~ and ~i~ ~1~, the equation would be:

$$\displaystyle {n \choose i} {m \choose i}$$

But since we want all possible ways to choose identical numbers of ~0~ and ~1~, we must sum from ~0~ to ~n~ (since ~n~ is the smaller of the two). Then, we need to subtract ~1~ since the empty subset should not be counted.

The answer is:

$$\displaystyle \sum_{i=0}^{n} {n \choose i} {m \choose i} = \sum_{i=0}^{n} {n \choose i} {m \choose m - i}$$

By Vandermonde's identity:

$$\displaystyle \sum_{i=0}^{n} {n \choose i} {m \choose m - i} = {n + m \choose m} = {n + m \choose n}$$

Subtracting ~1~:

$$\displaystyle {n + m \choose m} - 1$$

or

$$\displaystyle {n + m \choose n} - 1$$

How do we code the choose function?

$$\displaystyle {{n + m} \choose n} = \frac{(n+m)!}{n!m!}$$

But since numbers may be large, we just mod ~10^9+7~, which is a prime number. In modular arithmetic, we cannot divide, so we have to multiply the mod inverse instead.

By Fermat's Little Theorem:

$$\displaystyle a^{k-1} \equiv 1 (mod k)$$

$$\displaystyle a^{k-2} \equiv a^{-1} (mod j)$$

So we write the choose function as:

$$\displaystyle (n+m)! (n!)^{-1}(m!)^{-1}$$