<s>[INST] The probability that a binomial coefficient is divisible by 12. I do not know how to solve the following question:

What is the probability that $12$ divides $\binom{n}{12}$ for a randomly chosen natural number n?

Context:
I set this as a question in a problem set. I was trying to write these problems from memory while setting the questions. 
Anyway, now I am unable to solve this variant. Essentially $12^5$ divides $12!$ and thus the question reduces to

Find the values of natural $n$ for which $12^6$ divides $n(n-1)(n-2)\cdots (n-11)$. 

We can see that $12^5$ divides product of any consecutive $12$ numbers, so we need an additional $12$ from the product. When does this happen?
 [/INST] $$\binom{n}{12}=\frac{n(n-1)(n-2)\ldots(n-11)}{12!}=\frac{n(n-1)(n-2)\ldots(n-11)}{2^{10}\cdot 3^5\cdot 5^2\cdot 7\cdot 11}$$
Any $12$ consecutive integers include $4$ consecutive factors of $3$, at least one of which must be a multiple of $9$, and six consecutive multiples of $2$. Three of these must be multiples of $4$, and at least one must be a multiple of $8$, so their product has at least $10$ factors of $2$. We want to know when the product has at least $6$ factors of $3$ and at least $12$ factors of $2$.
The $12$ integers include two multiples of $9$ if $n$ is congruent to $0,1$, or $2$ modulo $9$, and they include a multiple of $27$ if $n$ is congruent to $0,1,2,\ldots,10$, or $11$ modulo $27$; thus, their product has at least $6$ factors of $3$ if and only if $n$ is congruent to one of the following $15$ numbers modulo $27$:
$$0,1,2,3,4,5,6,7,8,9,10,11,18,19,20$$
If the $12$ integers include two multiples of $8$, one of those must be a multiple of $16$, and their product will then have at least $12$ factors of $2$. If they include only one multiple of $8$, it must be a multiple of $32$ if the product is to have at least $12$ factors of $2$. They include two multiples of $8$ if $n\equiv 0,1,2,3\pmod8$, and they include a multiple of $32$ if $n$ is congruent to $0$, $1$, $2$, $3$, $4$, $5$, $6$, $7$, $8$, $9$, $10$, or $11$ modulo $32$. Thus, their product has at least $12$ factors of $2$ if and only if $n$ is congruent to one of the following $20$ numbers modulo $32$:
$$0,1,2,3,4,5,6,7,8,9,10,11,16,17,18,19,24,25,26,27$$
$\binom{n}{12}$ is a multiple of $12$ if and only if $n$ satisfies both of these conditions, meaning that it is in one of $15\cdot20=300$ residue classes modulo $27\cdot32=864$. As noted in the comments, the expression randomly chosen natural number is at best ill-defined, but we can at least say that the natural (or asymptotic) density of the set of natural numbers $n$ such that $12\mid\binom{n}{12}$ is 
$$\frac{300}{864}=\frac{25}{72}\;,$$
and this corresponds reasonably well to the intuitive notion.
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