As others have been saying, the answer numerically should be (m+n-2) choose (m-1) (mod 10^9 + 7). Of course, the problem is the time limit, which you will see if you try this method. Here are the two insights that I needed to solve this:

1. since the answer is mod p, make your own factorial function that takes mod p for every successive multiplication
2. Use FLT and fast exponentiation mod p to divide by (m-1)! and (n-1)! instead of calculating them and then dividing

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For those interested, the technique needed to solve this exercises is close to the implementation of math.comb in the Python standard library (albeit using modulo 2⁶⁴ which comes for free if using unsigned integers). You can watch Raymond Hettinger's keynote "Numerical Marvels Inside Python" for more explanations.

TLE'd on this but i was thinking somewhere along the lines of the followng: nc = int(math.factorial(m)//(math.factorial(r)*math.factorial(m-r))) where r in range(n)

My python3 solution but time limit execution

def solve(n):
    if n == 0:
        return 1
    i = 5
    while True:
        if (n*"0") in str(math.factorial(i)):
            return i
        i += 1