Solution: 2017 Winter Final - 3

Author: Michiel Smid

Question

Let $B$ be a set consisting of 45 bottles. Out of these, 17 are beer bottles, and the remaining 28 are cider bottles. Consider subsets of $B$ that contain

exactly 5 beer bottles and zero or more cider bottles,

exactly 5 cider bottles and zero or more beer bottles.

How many such subsets are there?

(a)

$ 2^{45} - {17 \choose 5} - {28 \choose 5} $

(b)

$ {17 \choose 5} \cdot 2^{28} + 2^{17} \cdot {28 \choose 5} - {17 \choose 5} \cdot {28 \choose 5} $

(c)

$ 2^{45} - {17 \choose 5} \cdot {28 \choose 5} $

(d)

$ {17 \choose 5} \cdot 2^{28} + 2^{17} \cdot {28 \choose 5} $

Solution

Holy moly, that’s a lot of words

Let’s do this anyway$ ( $cries internally $ ) $

Let B be the event that we have exactly 5 beer bottles and zero or more cider bottles
First, we choose 5 beer bottles out of the 17 beer bottles: $ \binom{17}{5} $
Then, choose how many cider bottles we want. We subsets with 0 cider bottles to 28 cider bottles
Somewhere out there, there's a rule that states that it's $ 2^{28} $
$ |B| = \binom{17}{5} \cdot 2^{28} $
Let C be the event that we have exactly 5 cider bottles and zero or more beer bottles
First, we choose 5 cider bottles out of the 28 cider bottles: $ \binom{28}{5} $
Then, choose how many beer bottles we want. We subsets with 0 beer bottles to 17 beer bottles
Somewhere out there, there's a rule that states that it's $ 2^{17} $
$ |C| = \binom{28}{5} \cdot 2^{17} $
Determine $ B \cap C $
We have exactly 5 beer bottles and 5 cider bottles
$ | B \cap C | = \binom{17}{5} \cdot \binom{28}{5} $

$ B \cup C = |B| + |C| - |B \cap C| $

$ B \cup C = \binom{17}{5} \cdot 2^{28} + \binom{28}{5} \cdot 2^{17} - \binom{17}{5} \cdot \binom{28}{5} $