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:
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 $n$