Not really sure what you're asking.

You might think about the Choose(n,k) function like this.

If order matters you have $n!$ permutations of $n$ items. That should be pretty clear.

Now when order doesn't matter you have permutations that are indistinguishable from one another.

A group of $n$ consists of the $k$ chosen, and the $n-k$ that aren't chosen.

For a given group of $k$ there are $k!$ ways those $k$ people can be arranged.

Similarly for the $n-k$ people not chosen, there are $(n-k)!$ ways they can be arranged.

So a given group of $k$ chosen people out of $n$ has $k! \times (n-k)!$ ways being arranged.

So if there are $n!$ total ways of being arranged and a choice of $k$ people can be arranged in $k! (n-k)!$ ways there are

$\dfrac {n!}{k!(n-k)!}$ ways to choose those $k$ people from the total $n$.

You did (a) just fine.

(b) should be pretty obvious. If 11 take part out of 23, how many don't take part? So how many ways can you choose those that don't take part out of 23?

Does this agree with your intuition?