What is the probability that $^{\text{n}}\text{C}_{7}$ is divisible by $12$?

Question

What is the probability that $^{\text{n}}\text{C}_{7}$ is divisible by $12$? Where $n$ can be take the values of all the natural numbers.

The number should be divisible by both $3$ and $4$ for this purpose. What I could do is individually find out which were divisible by 3 and which by 4 but wasn't able to find out the disjoint of those sets.

This was on my test, and the solution given is this, which I don't get a thing of:

You cannot speak of a probability here (unless some distribution on $\mathbb N$ is given and $n$ is a random number having that distribution). There is no canonical distribution (like the uniform on a finite set) on $\mathbb N$. — drhab, Apr 29 '19 at 11:57
I’m curious as to what you have for $Pr(3|{n \choose 7})$ and $Pr(4|{n \choose 7})$... — 雨が好きな人, Apr 29 '19 at 12:00
The probability is either $0$ or $1$, depending on the value of $n$. — Servaes, Apr 29 '19 at 12:02

score 2 · Answer 1 · answered Apr 29 '19 at 12:05

Look that $\binom{n}{7}=\frac{n!}{7!(n-7)!}$, and $12^2||7!$, so you need to look for an $n$ such that $\frac{n!}{(n-7)!}$ is at least a multiple of $12^3$, and it is easy to see that it occurs for $n\geq 12$. Now as the possible range for $n$ is $7\leq n \leq \infty$, so the numbers which do not satisfy $n$ are $7,8,9,10,11$. So as pointed out by Servaes, the probablity is equal to $0,1$ depending on the value of $n$

What is the probability that $^{\text{n}}\text{C}_{7}$ is divisible by $12$?

1 Answers1