Adventures of a Hole Punch

Firstly, we note that

$\dbinom{n}{7} = \dfrac{n(n-1)(n-2)(n-3)(n-4)(n-5)(n-6)}{2^4 \cdot 3^2 \cdot 5 \cdot 7}$

In order for this number to be divisible by $12=2^2 \cdot 3$ , the numerator must be divisible by $2^6 \cdot 3^3$ . (The 5 and 7 are ignored because by the Pigeonhole Principle, these will be cancelled out by factors in the numerator anyway). Thus, we wish to find all $n$ such that

$2^6 \cdot 3^3 \mid n(n-1)(n-2)(n-3)(n-4)(n-5)(n-6)$

Let's start by looking at the factors of 3.

By Pigeonhole Principle, 2 of the expressions in the product must be divisible by 3, so the whole expression is already divisible by 9. If $n\equiv 7, 8 \pmod{9}$ , then none of the brackets are divisible by 9, so the expression would not be divisible by 27. Thus, $n \equiv 0, 1, 2, 3, 4, 5, 6 \pmod{9}$ .

Let's look at the factors of 2 now.

If $n$ is even, we have that $n-2$ , $n-4$ and $n-6$ are also even, with two of them divisible by 4 by Pigeonhole Principle. Thus, this expression is already divisible by 64.

If $n$ is odd, we have $n-1$ , $n-3$ and $n-5$ to be even.

If $n-3$ is the only number divisible by 4, the whole expression will be divisible by 16. To ensure that the product is divisible by 64, we must have $n-3$ to be divisible by 16, so $n \equiv 3 \pmod{16}$ must also to be true.

If $n-1$ and $n-5$ are both divisible by 4, the whole expression is divisible by 32. To ensure the product is divisible by 64, we must have one of $n-1$ and $n-5$ to be divisible by 8. Thus, $n \equiv 1, 5 \pmod{8} \implies n \equiv 1, 5, 9, 13 \pmod{16}$ .

Combining all this information together, we get that $\dbinom{n}{7}$ is divisible by 12 if and only if $n$ satisfies all of the below criteria:

$\begin{cases} n &\equiv 0, 1, 2, 3, 4, 5, 6 \pmod{9}\\ n&\equiv 0, 1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 13, 14 \pmod{16} \end{cases}$

There are 7 possibilities $\pmod{9}$ and 13 possibilities $\pmod{16}$ , so there are $7 \times 13 = 91$ possibilities $\pmod{144}$ . Thus, as more numbers $n$ are checked, the probability that $\dbinom{n}{7}$ is divisible by 12 approaches $\boxed{\dfrac{91}{144}}$ . Thus, $m+n=235$ .