Question 1: Let $k,n$ be integers with $2\leq k\leq n$, consider set $S=\{1,2,...,n\}$. What is the number of $k$-element subsets of $S$ that do not contain $1$ and do not contain $2$?
A) $\binom{n-1}{k-1}$
B) $\binom{n-1}{k}$
C) $\binom{n-2}{k-2}$
D) $\binom{n-2}{k}$
Attempt: Kind of lost on how to attempt this so I took a shot in the dark.
I took $n=3$, so $S=\{1,2,3\}$ and found all $8$ subsets of $S$. I found that only $2$ subsets have no $1$ and $2$ (empty set and $\{3\}$). So I'm assuming $k=2$ and taking $n=2$ satisfies the option $A$ being $2$? The answer is option $D$ though.
Question 2: Let $k,n$ be integers with $2\leq k \leq n$, consider set $S={1,2,...,n}$. What is the number of $k$-element subsets of $S$ that do not contain $1$ or do not contain $2$?
A) $\binom{n-1}{k-1}+\binom{n-1}{k-1}$
B) $\binom{n-2}{k}$
C) $\binom{n}{k}-\binom{n-2}{k-2}$
D) $\binom{n}{k}-\binom{n-1}{k-1}-\binom{n-1}{k-1}$
Attempt: I tried doing the same thing as Q1. I got $4$ subsets that don't have $1$ and $4$ subsets that don't have $2$ using $n=3$. I got $8$ total? But that didn't make any sense.
I'm pretty lost on how to approach these logically. I can't grasp the k-element stuff
For 1, the answer should be D. If you exclude $1$ and $2$ you have $n-2$ elements left of $S$. You have to select $k$ of them. There are $n-2 \choose k$ ways to do that.
For 2, the same logic as in 1 says there are ${n-1 \choose k}$ subsets that do not contain $1$ and the same ${n-1 \choose k}$ that do not contain $2$, but we have counted the ones that contain neither twice, so we have to subtract them once and have $2{n-1 \choose k}-{n-2 \choose k}$. This is the inclusion-exclusion principle.