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?
 A: $\binom{6}{4}$ 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 $$\binom{6}{4}\cdot b^4 \cdot c^3. $$
A: Elisa must order a cider, thus we need to choose $4$ out of her friends to order beer. Which gives $\binom{6}{4} $ choices. For each of those $\binom{6}{4} $ choices, we must decide on the type of beer. Each choice implies $4$ orders of beer and $3$ orders of cider, which is where the answer came from.
Hence, to address your questions specifically, we have $\binom{6}{4} $ because we can choose which $4$ of her $6$ friends order beer. In addition, no matter which $4$ friends we choose to buy beer, we still have c choices of cider for the other $2$ friends, and Elisa of course, can have c choices of cider. The main thing is, we have already reserved a cider, but we didn't reserve a specific cider for her. She still has $c$ choices nevertheless.
