After the i-th iteration of the for loop, ans will contain n!/((n-i)!i!) which is exactly \binom{n}{i}, an integer.
Technically "ans" can grow above the final result in my example, but even that could be fixed if one really wants (e.g. i must divide either ans or n-i, you play a bit with divmod to figure out which division you do first.)
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It always divides it evenly, that's why it works.
After the i-th iteration of the for loop, ans will contain n!/((n-i)!i!) which is exactly \binom{n}{i}, an integer.
Technically "ans" can grow above the final result in my example, but even that could be fixed if one really wants (e.g. i must divide either ans or n-i, you play a bit with divmod to figure out which division you do first.)