Ways to people can get exactly 4 beers 
Let $b ≥ 1$ and $c ≥ 1$ be integers. Elisa’s neighborhood pub serves $b$ different types of
  beer and $c$ different types of cider. Elisa invites $6$ friends to this pub and orders $7$ drinks,
  one drink (beer or cider) for each friend, and one cider for herself. Different people may get
  the same type of beer or cider.
  In how many ways can Elisa place these orders, such that exactly $4$ people get a beer?
Answer: ${6\choose4} \cdot b^4 \cdot c^3$

Why do we do ${6\choose4}$ here? I can understand that $b^4$ is the number of ways you can distribute the beers to $4$ people, but if we then do $c^3$ thats $3$ ways to distribute $3$ ciders (including Elisa) which makes for $7$ people. Also why do we do $c^3$ when we have already reserved one cider for Elisa?

$\binom{6}{4}$ is the number of ways to choose the 4 of Elisa's friends who will get a beer. 
$b^4$ is the number of ways those 4 friends can choose their beers: Each of those friends can choose one of $b$ different beers, so there are $b^4$ different assignments of friends to beers. 
$c^3$ is the number of ways Elisa and the two cider-getting friends can get their drinks, since c is the number of choices of ciders that each person who receives cider has.
Putting this all together gives you $$\binom{6}{4}\cdot b^4 \cdot c^3. $$
