No. A multiplicative recurrence preserves integrality and avoids factorial-sized intermediate work.

Before writing a formula, describe one outcome and ask what changes make it a different outcome. The algebra comes after that identity decision.

What the function promises: Return C(n,k) for integers n>=0. Invalid k returns zero.

def choose(n, k):
    if n < 0: raise ValueError("Negative n")
    if k < 0 or k > n: return 0
    k = min(k, n-k)
    answer = 1
    for i in range(1, k+1):
        answer = answer * (n-k+i) // i
    return answer

assert choose(5,2) == 10
assert choose(100,0) == 1
assert choose(4,7) == 0

Work and storage: O(min(k,n−k)) arithmetic steps and O(1) stored integers; big-integer cost depends on result size.

When to recognize it: A multiplicative recurrence preserves integrality and avoids factorial-sized intermediate work.

Check before applying it: A memorized formula fails when objects repeat or constraints couple choices.

Modular inverses may not exist; use prime-factor exponents, CRT, or modulus-specific methods.