Let $n>3$ and $A=\{1, 2, \ldots , n\}$. How many subsets $B$ of $A$ contain both $1$ and $2$

Question

I think I was able to find the answer from some guesswork, but I'm unsure of how to prove the result.

If $n=5$, then there are $2^5=32$ total subsets by the power set rule.

The subsets that contain $1$ and $2$ are $\{1,2\}$, $\{1,2,3\}$, $\{1,2,3,4\}$, $\{1,2,3,4,5\}$, $\{1,2,4\}$, $\{1,2,4,5\}$, $\{1,2,5\}$, $\{1,2,3,5\}$.

There are $8$ subsets listed here, and I know that $8=2^3$. I also noticed that $2^3=2^{5-2}$.

Therefore, for any $n \in \mathbb{N}$, there would be $2^{n-2}$ subsets that contain $1$ and $2$.

For a set with $n$ elements, there would be $2^n$ subsets. To prove my formula rigorously, I know that I cannot just list out the subsets, so now I'm stuck.

Note: my question might be similar to this one.

Answer 1

You just have to ''add'' some subset of $\{3,4,5...,n\}$ to the set $\{1,2\}$. But the number of such subsets is exactly $2^{n-2}$ and you are done.